Showing posts with label logarithm. Show all posts
Showing posts with label logarithm. Show all posts

Thursday, July 1, 2021

Logarithm: Miscellaneous Exercise


Definition: Logarithm


Let $N$ and $b$ be positive real numbers, with $b \neq 1$. Then the logarithm of $N$ (with respect) to the base $b$ is the exponent by which $b$ must be raised to yield $N$, and is denoted by $\log _{b} N$

Rules of Logarithms


$\begin{array}{ll} \text{L}1. & N=b^{\log _{b} N}\\\\ \text{L}2. & x=\log _{b} b^{x}\\\\ \text{L}3. & \log _{b} b=1\\\\ \text{L}4. & \log _{b} 1=0\\\\ \text{L}5. & \log _{b}(M N)=\log _{b} M+\log _{b} N\\\\ \text{L}6. & \log _{b} N^{p}=p \log _{b} N\\\\ \text{L}7. & \log _{b}\left(\displaystyle\frac{M}{N}\right)=\log _{b} M-\log _{b} N\\\\ \text{L}8. & \log _{a} N=\displaystyle\frac{\log _{b} N}{\log _{b} N}\\\\ \text{L}9. & \log _{a} N=\displaystyle\frac{1}{\log _{N} a}\\\\ \text{L}10. & \log _{a^{p}} N=\displaystyle\frac{1}{p} \log _{a} N\\\\ \text{L}11. & a^{\log _{k} b}=b^{\log _{k} a} \end{array}$

Common Logarithm


The logarithm of $N$ to the base $10\left(\log _{10} N\right)$ is said to be a common logarithm, and is usually written as $\log N$ (omitting the base). where $n$ is called the characteristic and $\log a$ is called the mantissa of $\log N$.

$\begin{array}{|l|} \hline\log _{10} N=\log N\\ \hline \end{array}$


If $\quad N=a \times 10^{n}$,

then $\quad \log N=\log \left(a \times 10^{n}\right)=\log 10^{n}+\log a=n+\log a$ where $n$ is called the characteristic and $\log a$ is called the mantissa of $\log N$.

Note that $n$ is an integer and $1 \leq a<10$.

Euler's Number


As a positive integer $n$ become very large, the value of $\left(1+\displaystyle\frac{1}{n}\right)^{n}$ approaches an irrational number, which is denoted by $e$.

Natural Logarithm


The logarithm of $N$ to the base $e$ is called a natural logarithm, and is denoted by $\ln N$.

$\begin{array}{|l|} \hline\log _{e} N=\ln N\\ \hline \end{array}$

  1. Write the following expressions in terms of $\log x$, $\log y$ and $\log z$.
  2. (a) $\log (x^{2} y)$ Show Solution
    (b) $\log \displaystyle\frac{x^{3} y^{2}}{z}$ Show Solution
    (c) $\log \displaystyle\frac{\sqrt{x} \sqrt[3]{y^{2}}}{z^{4}}$ Show Solution
    (d) $\log (x y z)$ Show Solution
    (e) $\log \left(\displaystyle\frac{x}{y z}\right)$ Show Solution
    (f) $\log \left(\displaystyle\frac{x}{y}\right)^{2}$ Show Solution
    (g) $\log \left(x y\right)^{\frac{1}{3}}$ Show Solution
    (h) $\log (x \sqrt{z})$ Show Solution
    (i) $\log \displaystyle\frac{\sqrt[3]{x}}{\sqrt[3]{y z}}$ Show Solution
    (j) $\log \sqrt[4]{\displaystyle\frac{x^{3} y^{2}}{z^{4}}}$ Show Solution
    (k) $\log \left(x \sqrt{\displaystyle\frac{\sqrt{x}}{z}}\right)$ Show Solution
    (l) $\log \sqrt{\displaystyle\frac{x y^{2}}{z^{8}}}$ Show Solution

  3. Prove the following statements.
  4. (a) $\quad\log _{\sqrt{b}} x=2 \log _{b} x$ Show Solution
    (b) $\quad\log _{\frac{1}{\sqrt{b}}} \sqrt{x}=-\log _{b} x$ Show Solution
    (c) $\quad\log _{b^{4}} x^{2}=\log _{b} \sqrt{x}$ Show Solution

  5. Given that $\log 2=x, \log 3=y$ and $\log 7=z$, express the following expressions in terms of $x, y$, and $z$.
  6. (a) $\log 12$ Show Solution
    (b) $\log 200$ Show Solution
    (c) $\log \displaystyle\frac{14}{3}$ Show Solution
    (d) $\log 0.3$ Show Solution
    (e) $\log 1.5$ Show Solution
    (f) $\log 10.5$ Show Solution
    (g) $\log 15$ Show Solution
    (h) $\log \displaystyle\frac{6000}{7}$ Show Solution

  7. Solve the following logarithmic equations.

  8. (a) $\ln x=-3$ Show Solution
    (b) $\log (3 x-2)=2$ Show Solution
    (c) $2 \log x=\log 2+\log (3 x-4)$ Show Solution
    (d) $\log x+\log (x-1)=\log (4 x)$ Show Solution
    (e) $\log _{3}(x+25)-\log _{3}(x-1)=3$ Show Solution
    (f) $\log _{9}(x-5)+\log _{9}(x+3)=1$ Show Solution
    (g) $\log x+\log (x-3)=1$ Show Solution
    (h) $\log _{2}(x-2)+\log _{2}(x+1)=2$ Show Solution

  9. Find the inverse of each of the following functions.

  10. (a) $\quad f(x)=\log _{2}(x-3)-5$

    Show Solution

    (b) $\quad f(x)=3 \log _{3}(x+3)+1$

    Show Solution

    (c) $\quad f(x)=-2 \log 2(x-1)+2$

    Show Solution

    (d) $\quad f(x)=-\ln (1-2 x)+1$

    Show Solution

    (e) $\quad f(x)=2^{x}-3$

    Show Solution

    (f) $\quad f(x)=2 \cdot 3^{3 x}-1$

    Show Solution

    (g) $\quad f(x)=-5 \cdot e^{-x}+2$

    Show Solution

    (h) $\quad f(x)=1-2 e^{-2 x}$

    Show Solution

Friday, June 25, 2021

Multiple Choice : Logarithms

တက္ကသိုလ်ဝင်တန်း မေးခွန်းမှာ MCQ Format မပါတော့ပေမယ့် Multiple Choice Question ဆိုတာ ကျောင်းသားရဲ့ ဘာသာရပ်ဆိုင်ရာ နားလည်တတ်သိမှု၊ ဖြတ်ထိုးဉာဏ်၊ ဆင်ခြင်နိုင်စွမ်း စတာတွေကို စစ်ဆေးတာဖြစ်လို့ လေ့ကျင့်ထားသင့်ပါတယ်။ နိုင်ငံတကာ တက္ကသိုလ်ဝင် စာမေးပွဲများမှာလည်း MCQ ကိုပဲ ဦးစားပေး မေးလေ့ရှိတာမို့ နိုင်ငံရပ်ခြား ကျောင်းတက်ဖို့ စာမေးပွဲဖြေဆိုမည့်သူများ အတွက်လည်း အသုံးဝင်ပါလိမ့်မယ်။


Definition: Logarithm


Let $N$ and $b$ be positive real numbers, with $b \neq 1$. Then the logarithm of $N$ (with respect) to the base $b$ is the exponent by which $b$ must be raised to yield $N$, and is denoted by $\log _{b} N$

Rules of Logarithms


$\begin{array}{ll} \text{L}1. & N=b^{\log _{b} N}\\\\ \text{L}2. & x=\log _{b} b^{x}\\\\ \text{L}3. & \log _{b} b=1\\\\ \text{L}4. & \log _{b} 1=0\\\\ \text{L}5. & \log _{b}(M N)=\log _{b} M+\log _{b} N\\\\ \text{L}6. & \log _{b} N^{p}=p \log _{b} N\\\\ \text{L}7. & \log _{b}\left(\displaystyle\frac{M}{N}\right)=\log _{b} M-\log _{b} N\\\\ \text{L}8. & \log _{a} N=\displaystyle\frac{\log _{b} N}{\log _{b} N}\\\\ \text{L}9. & \log _{a} N=\displaystyle\frac{1}{\log _{N} a}\\\\ \text{L}10. & \log _{a^{p}} N=\displaystyle\frac{1}{p} \log _{a} N\\\\ \text{L}11. & a^{\log _{k} b}=b^{\log _{k} a} \end{array}$

Common Logarithm


The logarithm of $N$ to the base $10\left(\log _{10} N\right)$ is said to be a common logarithm, and is usually written as $\log N$ (omitting the base). where $n$ is called the characteristic and $\log a$ is called the mantissa of $\log N$.

$\begin{array}{|l|} \hline\log _{10} N=\log N\\ \hline \end{array}$

If $\quad N=a \times 10^{n}$,

then $\quad \log N=\log \left(a \times 10^{n}\right)=\log 10^{n}+\log a=n+\log a$ where $n$ is called the characteristic and $\log a$ is called the mantissa of $\log N$.

Note that $n$ is an integer and $1 \leq a<10$

Euler's Number


As a positive integer $n$ become very large, the value of $\left(1+\displaystyle\frac{1}{n}\right)^{n}$ approaches an irrational number, which is denoted by $e$.

Natural Logarithm


The logarithm of $N$ to the base $e$ is called a natural logarithm, and is denoted by $\ln N$.

$\begin{array}{|l|} \hline\log _{e} N=\ln N\\ \hline \end{array}$

MCQ Test


1. If $\log _{10} x=3$ then $x=$
A. $500$
B. $\displaystyle\frac{10}{3}$
C. $700$
D. $ 1000$
2. If $\log _{7} x=2$, then $x=$
A. $14$
B. $49$
C. $128$
D. $64$
3. The characteristic of log 19 is
A. $0$
B. $10$
C. $2$
D. $ 1$
4. The characteristic of $\log 3.216$ is
A. $0$
B. $4$
C. $3$
D. $10$
5. Common logarithm has the base
A. $2$
B. $e$
C. $\pi$
D. $ 10$
6. In scientific notation $0.00416$ is written as
A. $0.0416 \times 10^{-1}$
B. $0.416 \times 10^{-2}$
C. $ 4.16 \times 10^{-3}$
D. $41.6 \times 10^{-4}$
7. In decimal form $2.35 \times 10^{-2}$ is written as
A. $2.35$
B. $0.0235$
C. $0.00235$
D. $0.000235$
8. $\log 5+\log 8-\log 3=$
A. $5 \log \displaystyle\frac{8}{3}$
B. $3 \log 40$
C. $\log \displaystyle\frac{40}{3}$
D. $3 \log \displaystyle\frac{5}{8}$
9. $\log 50$ can be written as
A. $\log 2+2 \log 5$
B. $\log 2+\log 15$
C. $\log 2+5\log 2$
D. $\log 2+\log 5$
10. 3 is the characteristic in the logarithm of the number
A. $879.2$
B. $87.92$
C. $8.792$
D. $8792$
11. If $\log _{2} 8=x$ then $x=$
A. $2^{8}$
B. $64$
C. $3^{2}$
D. $3$
12. $5^{4}=625$ is written in the logarithmic form as
A. $\log 5=625$
B. $\log _{5} 4=625$
C. $\log _{5} 625=4$
D. $\log _{4} 625=5$
13. If $\log _{81} x=-\displaystyle\frac{3}{4}$ then $x=$
A. $27$
B. $\displaystyle\frac{1}{3}$
C. $\displaystyle\frac{1}{27}$
D. $\displaystyle\frac{1}{9}$
14. If antilog $3.8716=7440$ and $\log x=0.8716$ then $x=$
A. $74.40$
B. $7.440$
C. $744.0$
D. $7440$
15. If $\log 5=0.6990$ and $\log 3=0.4771$, then $\log 45=$
A. $1.6532$
B. $1.1761$
C. $1.8751$
D. $1.2219$
16. 3 log2 $-2$ log5 in the simplified form is
A. $\log \displaystyle\frac{6}{10}$
B. $\log \displaystyle\frac{9}{12}$
C. $\log \displaystyle\frac{8}{25}$
D. $\log \displaystyle\frac{25}{8}$
17. If $\log _{x} 81=4$ then $x=$
A. $3$
B. $2$
C. $-1$
D. $0$
18. If $\log _{8} x=\displaystyle\frac{2}{3}$ then $x=$
A. $2$
B. $4$
C. $3$
D. $-1$
19. $3 \log 2+\log 3=\log x$ then $x=$
A. $12$
B. $18$
C. $24$
D. $30$
20. If $\log _{4} 64=x$, then $x=$
A. $2$
B. $-1$
C. $0$
D. $3$
21. If $\log _{81} 9=x$, then $x=$
A. $3$
B. $2$
C. $1$
D. $\displaystyle\frac{1}{2}$
22. If $\log_{x} 49=2 ; x=$
A. $6$
B. $3$
C. $7$
D. $0$
23. If $\log 35+\log 36=\log (3 x)$ then $x=$
A. $400$
B. $420$
C. $520$
D. $600$
24. $\log 3+\log 6-\log 2=\log x$ then $x=$
A. $10$
B. $9$
C. $8$
D. $7$
25. $\log_{x} 36=2$ then $x=$
A. $5$
B. $8$
C. $2$
D. $6$
26. If $\log_{8} 16=x$, then $x=$
A. $\displaystyle\frac{4}{3}$
B. $\displaystyle\frac{1}{2}$,
C. $2$
D. $-\displaystyle\frac{1}{2}$
27. $\log 5+\log 8-\log 6=$
A. $\log 7$
B. $\log \displaystyle\frac{13}{6}$
C. $\log \displaystyle\frac{40}{6}$
D. $\log 50$
28. The characteristic of $\log 0.00329$ is
A. $\overline{1}$
B. $\overline{3}$
C. $\overline{2}$
D. $0$
29. The characteristic of $\log 1.02$ is
A. $1$
B. $\overline{3}$
C. $0$
D. $-1$
30. If $\log _{10} 100=x$, then $x=$
A. $2$
B. $1$
C. $0$
D. $-1$
31. The characteristic of $\log 0.000753$ is
A. $\overline{1}$
B. $\overline{2}$
C. $\overline{3}$
D. $\overline{4}$
32. The exponential form of $y=\log _{a} x$ is
A. $ x=y$
B. $a^{y}=x$
C. $x^{y}=a$
D. $a=x^{y}$
33. The logarithmic form of $a=y^x$ is
A. $\log _{a} x=y$
B. $ \log _{y} a=x$
C. $\log _{x} y=a$
D. $\log _{a} y=x$
34. The integral part of logarithm is called
A. determinant
B. matrix
C. mantissa
D. characteristic
35. The decimal part of logarithm is called
A. determinant
B. set
C. mantissa
D. characteristic
36. $ \displaystyle\frac{\log 5}{\log 3}=$
A. $\log 5-\log 3$
B. $\log_{3} 5$
C. $\log_{5} 3$
D. $\log\displaystyle\frac{ 5}{3}$
37. $\log 729=$
A. $\log 3$
B. $6 \log 3$
C. $\log 6$
D. $\log 3+\log 6$
38. If $\log 2=0.3010 . \log 3=0.4771, \log 5=0.6990$ then $\log 30=$
A. $1.4771$
B. $0.4771$
C. $-0.4771$
D. $-1.4771$
39. If $\log 2=0.3010, \log 3=0.4771$ then $\log 4.5=$
A. $0.7781$
B. $0.1761$
C.$0.6532$
D. $1.6532$
40. If $\log _{10} 7=a$, then $\log _{10}\left(\displaystyle\frac{1}{70}\right)=$
A. $-(1+a)$
B. $\displaystyle\frac{1}{1+a}$
C. $\displaystyle\frac{a}{10}$
D. $\displaystyle\frac{1}{10 a}$
41. $\displaystyle\frac{1}{\log _{a} b} \times \displaystyle\frac{1}{\log _{b} c} \times \displaystyle\frac{1}{\log _{c} a}=$
A. $-1$
B. $0$
C. $1$
D. $a b c$
42. If $\log _{10} 2=a$ and $\log _{10} 3=b$ then $\log _{5} 12=$
A. $\displaystyle\frac{a+b}{1+a}$
B. $\displaystyle\frac{2 a+b}{1+a}$
C. $\displaystyle\frac{a+2 b}{1+a}$
D. $\displaystyle\frac{2 a+b}{1-a}$
43. If $\log _{a}(a b)=x$, then $\log _{b}(a b)=$
A. $\displaystyle\frac{1}{x}$
B. $\displaystyle\frac{x}{x+1}$
C. $\displaystyle\frac{x}{1-x}$
D. $\displaystyle\frac{x}{x-1}$
44. $2 \log _{10} 5+\log _{10} 8-\displaystyle\frac{1}{2} \log _{10} 4=$
A. $2$
B. $4$
C. $2\left(1-\log _{10} 1\right)$
D. $4\left(1-\log _{10} 1\right)$
45. If $\log _{5}\left(x^{2}+x\right)-\log _{5}(x+1)=2$, then $x=$
A. $5$
B. $10$
C. $25$
D. $\displaystyle\frac{1}{5}$
46. If $\log _{10} x-5 \log _{10} 3=-2$, then $x=$
A. $\displaystyle\frac{80}{100}$
B. $\displaystyle\frac{81}{100}$
C. $\displaystyle\frac{125}{100}$
D. $\displaystyle\frac{243}{100}$
47. If $\log _{3} x+\log _{9} x^{2}+\log _{27} x^{3}=9$, then $x=$
A. $3$
B. $9$
C. $27$
D. $\displaystyle\frac{1}{3}$
48. If $a=\log _{8} 225$ and $b=\log _{2} 15$, then $\displaystyle\frac{a}{b}=$
A. $\displaystyle\frac{1}{3}$
B. $\displaystyle\frac{2}{3}$
C. $\displaystyle\frac{3}{2}$
D. $3$
49. If the logarithm of a number is $-3.153$, what are characteristic and mantissa?
A. characteristic $=-4, \quad$ mantissa $=0.847$
B. characteristic $=-4, \quad$ mantissa $=0.153$
C. characteristic $=4, \quad$ mantissa $=-0.847$
D. characteristic $=-3, \quad$ mantissa $=-0.153$
50. If $\log \left(\displaystyle\frac{a}{b}\right)+\log \left(\displaystyle\frac{b}{a}\right)=\log (a+b)$, then
A. $a+b=1$
B. $a-b=1$
C. $a=b$
D. $a^{2}+b^{2}=1$

Answer Keys


$\begin{array}{|ll|ll|ll|ll|ll|} \hline 1. &\text {D} & 2. &\text {B} & 3. &\text {D} & 4. &\text {A} & 5. &\text {D} \\ \hline 6. &\text {C} & 7. &\text {B} & 8. & \text{C} & 9. &\text {A }& 10.&\text {D} \\ \hline 11.&\text {D} & 12.&\text {C} & 13.&\text {C} & 14.&\text {B} & 15.&\text {A} \\ \hline 16.&\text {C} & 17.&\text {A} & 18.&\text {B} & 19.&\text {C} & 20.&\text {D} \\ \hline 21.&\text {D} & 22.&\text {C} & 23.&\text {B} & 24.&\text {B} & 25.&\text {D} \\ \hline 26.&\text {A} & 27.&\text {C} & 28.&\text {B} & 29.&\text {C} & 30.&\text {A} \\ \hline 31.&\text {D} & 32.&\text {B} & 33.&\text {B} & 34.&\text {D} & 35.&\text {C} \\ \hline 36.&\text {B} & 37.&\text {B} & 38.&\text {A} & 39.&\text {C} & 40.&\text {A} \\ \hline 41.&\text {C} & 42.&\text {D} & 43.&\text {D} & 44.& \text {A}& 45.&\text {C} \\ \hline 46.&\text {D} & 47.&\text {C} & 48.&\text {B} & 49.&\text {A} & 50.&\text {A} \\ \hline \end{array}$

Wednesday, July 29, 2020

Logarithms : Exercise (3.5) - Solutions

1.          Given that $\log 2.345=0.3701 .$ What are the characteristics and the mantissas of each of the followings?

(a) $\log 234,500$

(b) $\log 0.0002345$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ \ \ \ \log 2.345=0.3701\\\\\text{(a)}\ \ \ \log 234,500\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{5}}} \right)\\\\\ \ \ =\log {{10}^{5}}+\log 2.345\\\\\ \ \ =5+0.3701\\\\\ \ \ =5.3701\\\\\therefore \ \text{characteristic}=5\\\\\ \ \ \text{mantissa}=.3701\\\\\\\text{(b)}\ \ \ \log 0.0002345\\\\\ \ \ =\log \left( {2.345\times {{{10}}^{{-4}}}} \right)\\\\\ \ \ =\log {{10}^{{-4}}}+\log 2.345\\\\\ \ \ =-4+0.3701\\\\\therefore \ \text{characteristic}=-4\\\\\ \ \ \text{mantissa}=.3701\end{array}$

2.          Using $\log _{10} 2.74=0.4377, \log _{10} 2.83=0.4518, \log _{10} 5.97=0.7760$, $\log _{10} 6.21=0.7931, \log _{10} 8.18=0.9128$ and $\log _{10} 9.27=0.9671$, compute

(a) $\left(\displaystyle\frac{28.3}{597 \times 621}\right)^{2}$

(b) $\displaystyle\frac{274^{\frac{1}{3}}}{927 \times 818}$

(c) $\displaystyle\frac{28.3 \sqrt{0.621}}{597}$

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ {{\log }_{{10}}}2.74=0.4377,\\\ \ \ \ \ {{\log }_{{10}}}2.83=0.4518,\\\ \ \ \ \ {{\log }_{{10}}}5.97=0.7760,\\\ \ \ \ \ {{\log }_{{10}}}6.21=0.7931,\\\ \ \ \ \ {{\log }_{{10}}}8.18=0.9128,\\\ \ \ \ \ {{\log }_{{10}}}9.27=0.9671.\\\text{(a)}\ \ \text{Let}\ x={{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}},\ \text{then}\\\ \ \ \ \ \log x=\log {{\left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)}^{2}}\\\ \ \ \ \ \ \ \ \ \ \ \ =2\log \left( {\displaystyle\frac{{28.3}}{{597\times 621}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log (597\times 621)} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {\log 28.3-\log 597-\log 621} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =2\left( {1.4518-2.7760-2.7931} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =-8.2346\\\ \ \ \ \ x={{10}^{{-8.2346}}}\\\\\text{(b)}\ \ \text{Let}\ x=\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{{{{274}}^{{\displaystyle\frac{1}{3}}}}}}{{927\times 818}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\log \left( {927\times 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\log 274-\left( {\log 927+\log 818} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{3}\left( {2.4377} \right)-2.9671-2.9128\\\ \ \ \ \ \ \ \ \ \ \ \ =-5.0673\\\ \ \ \ \ x={{10}^{{-5.0673}}}\\\\\text{(c)}\ \ \text{Let}\ x=\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}},\ \text{then}\\\ \ \ \ \ \log x=\log \left( {\displaystyle\frac{{28.3\sqrt{{0.621}}}}{{597}}} \right)\\\ \ \ \ \ \ \ \ \ \ \ \ =\log 28.3+\displaystyle\frac{1}{2}\log 0.621-\log 597\\\ \ \ \ \ \ \ \ \ \ \ \ =1.4518+\displaystyle\frac{1}{2}\left( {-1+0.7931} \right)-2.7760\\\ \ \ \ \ \ \ \ \ \ \ \ =-1.4277\\\ \ \ \ \ x={{10}^{{-1.4277}}}\end{array}$

3.          Express each to a natural logarithm in terms of $\ln 2$ and $\ln 5$.

(a) $\log_{8}25$

(b) $\log_{5}64$

(c) $\log_{25}100$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ \ {{\log }_{8}}25\ \\\ \ \ \ =\displaystyle\frac{{\ln 25}}{{\ln 8}}\\\ \ \ \ =\displaystyle\frac{{\ln {{5}^{2}}}}{{\ln {{2}^{3}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5}}{{3\ln 2}}\\\\\text{(b)}\ \ \ \ {{\log }_{5}}64\ \\\ \ \ \ =\displaystyle\frac{{\ln 64}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{\ln {{2}^{6}}}}{{\ln 5}}\\\ \ \ \ =\displaystyle\frac{{6\ln 5}}{{\ln 2}}\\\\\text{(c)}\ \ \ \ {{\log }_{{25}}}100\ \\\ \ \ \ =\displaystyle\frac{{\ln 100}}{{\ln 25}}\\\ \ \ \ =\displaystyle\frac{{\ln ({{5}^{2}}\times {{2}^{2}})}}{{\ln {{5}^{2}}}}\\\ \ \ \ =\displaystyle\frac{{2\ln 5+2\ln 2}}{{2\ln 5}}\\\ \ =1+\displaystyle\frac{{\ln 2}}{{\ln 5}}\ \ \ \end{array}$

4.          Write each expression as a single logarithm.

(a) $3\ln \left( {t + 5} \right) - 4\ln t - 2\ln \left( {s - 1} \right)$

(b) $2\ln x + 5\ln y - \displaystyle\frac{1}{2}\ln z$

(c) $\displaystyle \frac{1}{3}\ln a - 6\ln b + 2$

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$\begin{array}{l} \text{(a)}\ \ \ \ 3\ln \left( {t+5} \right)-4\ln t-2\ln \left( {s-1} \right)\\\\ \ \ \ \ =\ln {{\left( {t+5} \right)}^{3}}-\ln {{t}^{4}}-\ln {{\left( {s-1} \right)}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{{\left( {t+5} \right)}}^{3}}}}{{{{t}^{4}}{{{\left( {s-1} \right)}}^{2}}}}\\\\\\ \text{(b)}\ \ \ \ 2\ln x+5\ln y-\displaystyle\frac{1}{2}\ln z\\\\ \ \ \ \ =\ln {{x}^{2}}+\ln {{y}^{5}}-\ln \sqrt{z}\\\\ \ \ \ \ =\ln \displaystyle\frac{{{{x}^{2}}\cdot {{y}^{5}}}}{{\sqrt{z}}}\\\\\\ \text{(c)}\ \ \ \ \displaystyle\frac{1}{3}\ln a-6\ln b+2\\\\ \ \ \ \ =\ln {{a}^{{\frac{1}{3}}}}-\ln {{b}^{6}}+\ln {{e}^{2}}\\\\ \ \ \ \ =\ln \displaystyle\frac{{\sqrt[3]{a}\cdot {{e}^{2}}}}{{{{b}^{6}}}} \end{array}$

5.          Solve the following logarithmic equations for $x$.

(a) $\ln \left( x \right) + \ln \left( {x + 3} \right) = \ln \left( {20 - 5x} \right)$

(b) $2\ln \left( {\sqrt x } \right) - \ln \left( {1 - x} \right) = 2$

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$\begin{array}{l}\text{(a)}\ \ \ln (x)+\ln \left( {x+3} \right)=\ln \left( {20-5x} \right)\\\ \\\ \ \ \ \ \ \ln \left( {x\left( {x+3} \right)} \right)=\ln \left( {20-5x} \right)\\\\\ \ \ \ \ \ x\left( {x+3} \right)=20-5x\\\\\ \ \ \ \ \ {{x}^{2}}+8x-20=0\\\\\ \ \ \ \ \ (x+10)(x-2)=0\\\\\ \ \ \ \ \ x=-10\ \text{or}\ x=2\\\\\ \ \ \ \ \ \text{Since }x>0,\ x=-10\ \text{is impossible}\text{.}\\\\\ \ \ \therefore \ \ x=2\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 2\ln \left( {\sqrt{x}} \right)-\ln \left( {1-x} \right)=2\\\ \\\ \ \ \ \ \ln \left( {\displaystyle\frac{{{{{\left( {\sqrt{x}} \right)}}^{2}}}}{{1-x}}} \right)=2\\\\\ \ \ \ \ \displaystyle\frac{x}{{1-x}}={{e}^{2}}\\\ \\\ \ \ \ \ x={{e}^{2}}-{{e}^{2}}x\\\\\ \ \ \ \ {{e}^{2}}x+x={{e}^{2}}\\\\\ \ \ \ \ x\left( {{{e}^{2}}+1} \right)={{e}^{2}}\\\\\ \ \ \ \ x=\displaystyle\frac{{{{e}^{2}}}}{{{{e}^{2}}+1}}\end{array}$

6.          Solve the equations:

(a) $x - xe^{5x + 2} = 0$

(b) $7 + 15e^{1 - 3x} = 10$ using $\ln 5 = 1.6094$.

(c) $4e^{1 + 3x} - 9e^{5 - 2x} = 0$ using $\ln 2=0.6931$ and $\ln 3=1.0986$.

No(3) မှ No(6) အထိ မေးခွန်းများသည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ Natural Logarithm ဆိုင်ရာ ပုစ္တာများကို တိုးချဲ့လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

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$\begin{array}{l}\text{(a)}\ \ \ x-x{{e}^{{5x+2}}}=0\\\ \\\ \ \ \ \ \ x\left( {1-{{e}^{{5x+2}}}} \right)=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ 1-{{e}^{{5x+2}}}=0\\\\\ \ \ \ \ \ x=0\ \text{or}\ {{e}^{{5x+2}}}=1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=\ln 1\\\\\ \ \ \ \ \ x=0\ \text{or}\ 5x+2=0\\\\\ \ \therefore \ \ \ x=0\ \text{or}\ x=-\displaystyle\frac{2}{5}\\\\\ \ \ \ \ \\\text{(b)}\ \ \ 7+15{{e}^{{1-3x}}}=10\\\ \\\ \ \ \ \ \ 15{{e}^{{1-3x}}}=3\\\\\ \ \ \ \ \ {{e}^{{1-3x}}}=\displaystyle\frac{1}{5}\\\ \\\ \ \ \ \ \ 1-3x=\ln \left( {\displaystyle\frac{1}{5}} \right)\\\\\ \ \ \ \ \ 1-3x=\ln \left( {{{5}^{{-1}}}} \right)\\\\\ \ \ \ \ \ 1-3x=-\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\ln \left( 5 \right)\\\\\ \ \ \ \ \ 3x=1+\text{1}\text{.6094}\\\\\ \ \ \ \ \ 3x=2.\text{6094}\\\\\ \ \ \ \ \ x=\text{0}\text{.8698}\\\\\\\text{(c)}\ \ \ 4{{e}^{{1+3x}}}-9{{e}^{{5-2x}}}=0\\\\\ \ \ \ \ 4{{e}^{{1+3x}}}=9{{e}^{{5-2x}}}\\\\\ \ \ \ \ \displaystyle\frac{{{{e}^{{1+3x}}}}}{{{{e}^{{5-2x}}}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ {{e}^{{5x-4}}}=\displaystyle\frac{9}{4}\\\\\ \ \ \ \ 5x-4=\ln \displaystyle\frac{{{{3}^{2}}}}{{{{2}^{2}}}}\\\\\ \ \ \ \ 5x=2\left( {\ln 3-\ln 2} \right)+4\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\ln 3-\ln 2+2} \right)\\\\\ \ \ \ \ x=\displaystyle\frac{2}{5}\left( {\text{1}\text{.0986}-\text{0}\text{.6931+2}} \right)\\\\\ \ \ \ \ x=\text{0}\text{.9622}\end{array}$

Tuesday, July 28, 2020

Logarithms : Exercise (3.4) - Solutions

1.          If $\log _{a} b+\log _{b} a^{2}=3,$ find $b$ in terms of $a$.

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$\begin{array}{l}{{\log }_{a}}\,b+{{\log }_{b}}\,{{a}^{2}}=3\\\\{{\log }_{a}}\,b+2{{\log }_{b}}a=3\\\\{{\log }_{a}}\,b+\displaystyle\frac{2}{{{{{\log }}_{a}}\,b}}=3\\\\\text{Let}\ {{\log }_{a}}\,b=u,\ \text{then}\\\\u+\displaystyle\frac{2}{u}=3\\\\\therefore \ \ {{u}^{2}}+2=3u\\\\\therefore \ \ {{u}^{2}}-3u+2=0\\\\\therefore \ \ (u-1)(u-2)=0\\\\\therefore \ \ {{\log }_{a}}\,b=1\ \text{or}\ {{\log }_{a}}\,b=2\\\\\therefore \ \ \,b=a\ \text{or}\ \,b={{a}^{2}}\end{array}$

2.          Show that

(a) $\log _{4} x=2 \log _{16} x$.

(b) $\log _{b} x=3 \log _{b^{3}} x \quad$.

(c) $\log _{2} x=\left(1+\log _{2} 3\right) \log _{6} x$.

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$\begin{array}{l}(\text{a})\ \ \ \text{LHS}={{\log }_{4}}x\\\\\ \ \ \ \ \ \text{RHS}=2{{\log }_{{16}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{{{4}^{2}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{2}{2}{{\log }_{4}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{4}}x\\\\\ \ \therefore \ \ {{\log }_{4}}x=2{{\log }_{{16}}}x\\\\\\(\text{b})\ \ \text{LHS}={{\log }_{b}}x\\\\\ \ \ \ \ \text{RHS}=3{{\log }_{{{{b}^{3}}}}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{3}{3}{{\log }_{b}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}x\\\\\ \ \therefore \ \ {{\log }_{b}}x=3{{\log }_{{{{b}^{3}}}}}x\\\\\\(\text{c})\ \ \text{LHS}={{\log }_{2}}x\\\\\ \ \ \ \ \text{RHS}=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {{{{\log }}_{2}}2+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}6\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{1}{{{{{\log }}_{6}}2}}\cdot {{\log }_{6}}x\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle\frac{{{{{\log }}_{6}}x}}{{{{{\log }}_{6}}2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{2}}x\\\\\ \ \therefore \ \ {{\log }_{2}}x=\left( {1+{{{\log }}_{2}}3} \right){{\log }_{6}}x\\\end{array}$

3.          If $a=\log _{b} c, b=\log _{c} a$ and $c=\log _{a} b,$ prove that $a b c=1$.

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$ \begin{array}{l}\ \ \ a={{\log }_{b}}c,\ b={{\log }_{c}}a,\ c={{\log }_{a}}b\\\\\therefore \ \ abc={{\log }_{b}}c\cdot \ {{\log }_{c}}a\cdot {{\log }_{a}}b\\\\\therefore \ \ abc=\displaystyle\frac{{\log c}}{{\log b}}\cdot \ \displaystyle\frac{{\log a}}{{\log c}}\cdot \displaystyle\frac{{\log b}}{{\log a}}\ \left[ {\because \ \text{by L8}} \right]\\\\\therefore \ \ abc=1\end{array}$

4.          Show that

(a) $\left(\log _{10} 4-\log _{10} 2\right) \log _{2} 10=1$.

(b) $2 \log _{2} 3\left(\log _{9} 2+\log _{9} 4\right)=3$.

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$\begin{array}{l}\text{(a)}\ \ \ \left( {{{{\log }}_{{10}}}4-{{{\log }}_{{10}}}2} \right)\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}\displaystyle\frac{4}{2}\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot {{\log }_{2}}10\\\\\ \ \ =\ {{\log }_{{10}}}2\cdot \displaystyle\frac{1}{{{{{\log }}_{{10}}}2}}\\\\\ \ \ =\ 1\\\\\\\text{(b)}\ \ \ 2\cdot {{\log }_{2}}3\cdot \left( {{{{\log }}_{9}}2+{{{\log }}_{9}}4} \right)\\\\\ \ \ ={{\log }_{2}}{{3}^{2}}\cdot {{\log }_{9}}(2\times 4)\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}8\\\\\ \ \ ={{\log }_{2}}9\cdot {{\log }_{9}}{{2}^{3}}\\\\\ \ \ ={{\log }_{2}}9\cdot 3{{\log }_{9}}2\\\\\ \ \ =3\cdot \displaystyle\frac{1}{{{{{\log }}_{9}}2}}\cdot {{\log }_{9}}2\\\\\ \ \ =3\end{array}$

5.          Compute

(a) $3^{\log _{2} 5}-5^{\log _{2} 3}$

(b) $4^{\log _{2} 3}$

(c) $2^{\log _{2 \sqrt{2}} 27}$

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$\begin{array}{l}\text{(a)}\ \ \ {{3}^{{{{{\log }}_{2}}\,5}}}-{{5}^{{{{{\log }}_{2}}\,3}}}\\\\\ \ \ \ ={{3}^{{{{{\log }}_{2}}\,5}}}-{{3}^{{{{{\log }}_{2}}\,5}}}\\\\\ \ \ \ =0\\\\\text{(b)}\ \ \ {{4}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{2}}} \right)}^{{{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\\\\\\\text{(c)}\ \ \ {{2}^{{{{{\log }}_{{2\sqrt{2}}}}27}}}\\\\\ \ \ \ ={{2}^{{{{{\log }}_{{{{2}^{{3/2}}}}}}{{3}^{3}}}}}\\\\\ \ \ \ ={{2}^{{\frac{2}{3}\times 3{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{2}^{{2{{{\log }}_{2}}3}}}\\\\\ \ \ \ ={{\left( {{{2}^{{{{{\log }}_{2}}3}}}} \right)}^{2}}\\\\\ \ \ \ ={{3}^{2}}\\\\\ \ \ \ =9\end{array}$

6.          Using properties of logarithm, solve the following equations.

(a) $ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2$

(b) $25^{\log _{10} x}=5+4 x^{\log _{10} 5}$

(c) $9^{\log _{3}\left(\log _{2} x\right)}=\log _{2} x-\left(\log _{2} x\right)^{2}+1$

(d) $\log _{2} x+\log _{4} x+\log _{16} x=7$

(e) $\log _{7} 2+\log _{49} x=\log _{\frac{1}{7}} \sqrt{3}$

(f) $\log _{3} x-\log _{\frac{1}{3}} x^{2}=6$

(g) $ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4$

No (6) သည် ပြဌာန်းစာအုပ်တွင် မပါဝင်ပါ။ Change of Base နှင့် သက်ဆိုင်သော ဉာဏ်စမ်းမေးခွန်း များကို ထပ်ဆင့်လေ့ကျင့်နိုင်ရန် ပေါင်းထည့်ပေးထားခြင်း ဖြစ်ပါသည်။

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$ \begin{array}{l} \text{(a)}\ \ {{\log }_{x}}10+{{\log }_{{{{x}^{2}}}}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10+\displaystyle\frac{1}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ \displaystyle\frac{3}{2}{{\log }_{x}}10=2\\\\ \ \ \ \ \ {{\log }_{x}}10=\displaystyle\frac{4}{3}\\\\ \ \ \ \ \ {{x}^{{\frac{4}{3}}}}=10\\\\ \ \ \ \ \ x={{10}^{{\frac{3}{4}}}}\\\\\\ \text{(b)}\ \ {{25}^{{{{{\log }}_{{10}}}x}}}=5+4{{x}^{{{{{\log }}_{{10}}}5}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{2}}} \right)}^{{{{{\log }}_{{10}}}x}}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{5}^{{2\cdot }}}^{{{{{\log }}_{{10}}}x}}=5+4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}\\\\ \ \ \ \ \ {{\left( {{{5}^{{{{{\log }}_{{10}}}x}}}} \right)}^{2}}-4\cdot {{5}^{{{{{\log }}_{{10}}}x}}}-5=0\\\\ \ \ \ \ \ \text{Let}\ {{5}^{{{{{\log }}_{{10}}}x}}}=u,\ \text{then}\\\\ \ \ \ \ \ {{u}^{2}}-4u-5=0\\\\ \ \ \ \ \ (u-5)(u+1)=0\\\\ \ \therefore \ \ \ u=5\ \text{or}\ u=-1\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\ \text{or}\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\\\\ \ \ \ \ \ \text{Since}\ {{5}^{{{{{\log }}_{{10}}}x}}}>0,\ {{5}^{{{{{\log }}_{{10}}}x}}}=-1\ \text{is}\ \text{impossible}\text{.}\\\\ \ \therefore \ \ \ {{5}^{{{{{\log }}_{{10}}}x}}}=5\\\\ \ \therefore \ \ \ {{\log }_{{10}}}x=1\\\\ \ \therefore \ \ \ x=10\\\\\\ \text{(c)}\ \ {{9}^{{{{{\log }}_{3}}\left( {{{{\log }}_{2}}x} \right)}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}9}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{{{{{\log }}_{3}}{{3}^{2}}}}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\\\ \ \ \ \ \ {{\left( {{{{\log }}_{2}}x} \right)}^{2}}={{\log }_{2}}x-{{\left( {{{{\log }}_{2}}x} \right)}^{2}}+1\\ \ \ \ \ \ 2{{\left( {{{{\log }}_{2}}x} \right)}^{2}}-{{\log }_{2}}x-1=0\\\\ \ \ \ \ \ \left( {2\cdot {{{\log }}_{2}}x+1} \right)\left( {{{{\log }}_{2}}x-1} \right)=0\\\\ \ \ \ \ \ {{\log }_{2}}x=-\displaystyle\frac{1}{2}\ \text{or}\ {{\log }_{2}}x=1\\\\ \ \therefore \ \ \ x={{2}^{{-\frac{1}{2}}}}\ \text{or}\ x=2\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{{\sqrt{2}}}\ \text{or}\ x=2\\\\\\ \text{(d)}\ \ {{\log }_{2}}x+{{\log }_{4}}x+{{\log }_{{16}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+{{\log }_{{{{2}^{2}}}}}x+{{\log }_{{{{2}^{4}}}}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x+\displaystyle\frac{1}{2}{{\log }_{2}}x+\displaystyle\frac{1}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ \displaystyle\frac{7}{4}{{\log }_{2}}x=7\\\\ \ \ \ \ \ {{\log }_{2}}x=4\\\ \therefore \ \ \ x={{2}^{4}}=16\\\\\\ \text{(e)}\ \ {{\log }_{7}}2+{{\log }_{{49}}}x={{\log }_{{\displaystyle\frac{1}{7}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{{{{7}^{2}}}}}x={{\log }_{{{{7}^{{-1}}}}}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+\displaystyle\frac{1}{2}{{\log }_{7}}x=-{{\log }_{7}}\sqrt{3}\\\\ \ \ \ \ \ {{\log }_{7}}2+{{\log }_{7}}{{x}^{{\frac{1}{2}}}}={{\log }_{7}}{{3}^{{-\frac{1}{2}}}}\\\\ \ \ \ \ \ {{\log }_{7}}2{{x}^{{\frac{1}{2}}}}={{\log }_{7}}\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 2{{x}^{{\frac{1}{2}}}}=\displaystyle\frac{1}{{{{3}^{{\frac{1}{2}}}}}}\\\\ \ \ \therefore \ \ 4x=\displaystyle\frac{1}{3}\\\ \ \therefore \ \ x=\displaystyle\frac{1}{{12}}\\\\\\ \text{(f)}\ \ {{\log }_{3}}x-{{\log }_{{\displaystyle\frac{1}{3}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x-{{\log }_{{{{3}^{{-1}}}}}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}x+{{\log }_{3}}{{x}^{2}}=6\\\\ \ \ \ \ \ {{\log }_{3}}{{x}^{3}}=6\\\\ \ \ \ \ \ {{x}^{3}}={{3}^{6}}\\\\ \ \therefore \ \ \ x={{3}^{2}}=9\\\\\\ \text{(g)}\ \ {{\log }_{x}}(9{{x}^{2}})\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ \displaystyle\frac{{{{{\log }}_{3}}(9{{x}^{2}})}}{{{{{\log }}_{3}}x}}\cdot {{\left( {{{{\log }}_{3}}x} \right)}^{2}}=4\\\\ \ \ \ \ \ {{\log }_{3}}(9{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}9+{{\log }_{3}}{{x}^{2}})\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ ({{\log }_{3}}{{3}^{2}}+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \ \ \ (2+2{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=4\\\\ \ \therefore \ \ (1+{{\log }_{3}}x)\cdot \left( {{{{\log }}_{3}}x} \right)=2\\\\ \ \therefore \ \ {{\left( {{{{\log }}_{3}}x} \right)}^{2}}+\left( {{{{\log }}_{3}}x} \right)-2=0\\\\ \ \therefore \ \ ({{\log }_{3}}x+2)({{\log }_{3}}x-1)=0\\\\ \ \therefore \ \ \ {{\log }_{3}}x=-2\ \text{or}\ {{\log }_{3}}x=1\\\\ \ \therefore \ \ \ x={{3}^{{-2}}}\ \text{or}\ x=3\\\\ \ \therefore \ \ \ x=\displaystyle\frac{1}{9}\ \text{or}\ x=3 \end{array}$

Friday, June 26, 2020

Logarithms : Exercise (3.3) Solutions



$\begin{array}{|l|l|l|} \hline {\ \ \ \ \ \ \ \ \ \ \text { Properties }} & {\ \ \ \ \ \ \ \text { For Exponents }} & {\ \ \ \ \ \ \ \ \ \ \ \ \ \text { For Logarithms }} \\ \hline \text { One-to-one Property } & \text { If } b^{x}=b^{y}, \text { then } x=y & \text { If } \log _{b} M=\log _{b} N, \text { then } M=N \\ \hline \text { Product Property } & b^{x} \cdot b^{y}=b^{x+y} & \log _{b}(M N)=\log _{b} M+\log _{b} N \\ \hline \text { Quotient Property } & \displaystyle\frac{b^{x}}{b^{y}}=b^{x-y} & \log _{b} \displaystyle\frac{M}{N}=\log _{b} M-\log _{b} N \\ \hline \text { Power Property } & \left(b^{x}\right)^{y}=b^{x y} & \log _{b} N^{p}=p \log _{b} N \\ \hline \end{array}$

1.           Replace $\square$ with the appropriate number.

              $\begin{array}{l} \text{(a)}\ \ \log _{3} 24=\log _{3} 6+\log _{3} \square\\ \text{(b)}\ \ \log _{5} 24=\log _{5} 60+\log _{5} \square\\ \text{(c)}\ \ \log _{2} \square=3 \log _{2} 3\\ \text{(d)}\ \ \log _{10} 9=\square \log _{10} 3\\ \text{(e)}\ \ \log _{8} 5=\log _{8} \square-\log _{8} 11 \end{array}$

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$\begin{array}{l} \text{(a)}\ \ 4\\ \text{(b)}\ \ \displaystyle\frac{2}{5}\\ \text{(c)}\ \ 27\\ \text{(d)}\ \ 2\\ \text{(e)}\ \ 55 \end{array}$

2.           Write each expression as a single logarithm.

              $\begin{array}{l} \text{(a)}\ \ \log _{b} 20+\log _{b} 57-\log _{b} 241\\ \text{(b)}\ \ 3 \log _{b} 8-\displaystyle\frac{1}{2} \log _{b} 12\\ \text{(c)}\ \ \log _{b} x-2 \log _{b} y-\log _{b} a\\ \text{(d)}\ \ \log _{2} 3+\log _{4} 15 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ {{\log }_{b}}20+{{\log }_{b}}57-{{\log }_{b}}241\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{20\times 57}}{{241}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{1140}}{{241}}\\\\ \text{(b)}\;\;\ \ 3{{\log }_{b}}8-\displaystyle \frac{1}{2}{{\log }_{b}}12\\ \ \ \ \ =\ {{\log }_{b}}{{8}^{3}}-{{\log }_{b}}{{12}^{{\displaystyle \frac{1}{2}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{\sqrt{{12}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{2\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256}}{{\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256\sqrt{3}}}{3}\\\\ \text{(c)}\;\;\ \ {{\log }_{b}}x-2{{\log }_{b}}y-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}x-{{\log }_{b}}{{y}^{2}}-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{x}{{a{{y}^{2}}}}\\\\ \text{(d)}\;\;\ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ \text{Let}\ {{\log }_{4}}15=x,\ \text{then}\\ \ \ \ \ 15={{4}^{x}}\\\ \ \ \ 15={{2}^{2}}^{x}\\ \ \ \ \ \therefore \ \ 2x={{\log }_{2}}15\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{1}{2}{{\log }_{2}}15\\ \ \ \ \ \therefore \ \ {{\log }_{4}}15={{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ \therefore \ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ =\ \ {{\log }_{2}}3+{{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ =\ \ {{\log }_{2}}3\sqrt{{15}} \end{array}$

3.           Write each expression in terms of $\log _{b} 2, \log _{b} 3$ and $\log _{b} 5$.

              $\begin{array}{l} \text{(a)}\ \ \log _{b} 8\\ \text{(b)}\ \ \log _{b} 15\\ \text{(c)}\ \ \log _{b} 270\\ \text{(d)}\ \ \log _{b} \displaystyle\frac{27 \sqrt[3]{5}}{16}\\ \text{(e)}\ \ \log _{b} \displaystyle\frac{216}{\sqrt[3]{32}}\\ \text{(f)}\ \ \log _{b}(648 \sqrt{125})\\ \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{b}}8={{\log }_{b}}{{2}^{3}}=3{{\log }_{b}}2\\\\ \text{(b)}\;\;{{\log }_{b}}15={{\log }_{b}}(3\times 5)={{\log }_{b}}3+{{\log }_{b}}5\\\\ \text{(c)}\;\;{{\log }_{b}}270={{\log }_{b}}(2\times {{3}^{3}}\times 5)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}2+{{\log }_{b}}{{3}^{3}}+{{\log }_{b}}5\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}2+3{{\log }_{b}}3+{{\log }_{b}}5\\\\ \text{(d)}\;\;{{\log }_{b}}\displaystyle \frac{{27\sqrt[3]{5}}}{{16}}={{\log }_{b}}\displaystyle \frac{{{{3}^{3}}\times {{5}^{{\frac{1}{3}}}}}}{{{{2}^{4}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}{{3}^{3}}+{{\log }_{b}}{{5}^{{\frac{1}{3}}}}-{{\log }_{b}}{{2}^{4}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{b}}3+\displaystyle \frac{1}{3}{{\log }_{b}}5-4{{\log }_{b}}2\\\\ \text{(e)}\;\;{{\log }_{b}}\displaystyle \frac{{216}}{{\sqrt[3]{{32}}}}={{\log }_{b}}\displaystyle \frac{{{{2}^{3}}\times {{3}^{3}}}}{{{{2}^{{\frac{5}{3}}}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}\left( {{{2}^{{^{{\frac{4}{3}}}}}}\times {{3}^{3}}} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{4}{3}{{\log }_{b}}2+3{{\log }_{b}}3\\\\ \text{(f)}\;\;{{\log }_{b}}(648\sqrt{{125}})={{\log }_{b}}({{2}^{3}}\times {{3}^{4}}\times {{5}^{{\frac{3}{2}}}})\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}{{2}^{3}}+{{\log }_{b}}{{3}^{4}}+{{\log }_{b}}{{5}^{{\frac{3}{2}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{b}}2+4{{\log }_{b}}3+\displaystyle\frac{3}{2}{{\log }_{b}}5 \end{array}$

4.           Evaluate each expression.

              $\begin{array}{l} \text{(a)}\ \ \log _{2} 128\\ \text{(b)}\ \ \log _{3} 81^{4}\\ \text{(c)}\ \ \log _{\frac{1}{2}} 8\\ \text{(d)}\ \ \log _{8} 2\\ \text{(e)}\ \ \log _{3} \displaystyle\frac{\sqrt{3}}{81}\\ \text{(f)}\ \ \displaystyle\frac{\log _{3} \sqrt{3}}{\log _{3} 81}\\ \text{(g)}\ \ \displaystyle\frac{\log _{2} 25}{\log _{2} 5}\\ \text{(h)}\ \ \log _{4} 8 \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{2}}128={{\log }_{2}}{{2}^{7}}=7\\\\ \text{(b)}\;\;{{\log }_{3}}{{81}^{4}}={{\log }_{3}}{{\left( {{{3}^{4}}} \right)}^{4}}={{\log }_{3}}{{3}^{{16}}}=16\\\\ \text{(c)}\;\;{{\log }_{{\frac{1}{2}}}}8={{\log }_{{\frac{1}{2}}}}{{2}^{3}}={{\log }_{{\frac{1}{2}}}}{{\left( {\displaystyle\frac{1}{2}} \right)}^{{-3}}}=-3\\\\ \text{(d)}\;\;{{\log }_{8}}2={{\log }_{8}}{{8}^{{\frac{1}{3}}}}=\displaystyle \frac{1}{3}\\\\ \text{(e)}\;\;{{\log }_{3}}\displaystyle \frac{{\sqrt{3}}}{{81}}={{\log }_{3}}\displaystyle \frac{{{{3}^{{\frac{1}{2}}}}}}{{{{3}^{4}}}}={{\log }_{3}}{{3}^{{-\frac{7}{2}}}}=-\displaystyle \frac{7}{2}\\\\ \text{(f)}\;\;\displaystyle \frac{{{{{\log }}_{3}}\sqrt{3}}}{{{{{\log }}_{3}}81}}=\displaystyle \frac{{{{{\log }}_{3}}{{3}^{{\frac{1}{2}}}}}}{{{{{\log }}_{3}}{{3}^{4}}}}=\displaystyle \frac{{\frac{1}{2}}}{4}=\displaystyle \frac{1}{8}\\\\ \text{(g)}\;\;\displaystyle \frac{{{{{\log }}_{2}}25}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{{{{\log }}_{2}}{{5}^{2}}}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{2{{{\log }}_{2}}5}}{{{{{\log }}_{2}}5}}=2\\\\ \text{(h)}\;\;{{\log }_{4}}8={{\log }_{4}}\sqrt{{64}}={{\log }_{4}}{{4}^{{\frac{3}{2}}}}=\displaystyle \frac{3}{2} \end{array}$

5.           Use $\log _{10} 2=0.3010$ and $\log _{10} 3=0.4771$ to evaluate each of the following expressions.

              $\begin{array}{lll} \text{(a)}\ \ \log _{10} 6 & \text{(b)}\ \ \log _{10} 1.5 & \text{(c)}\ \ \log _{10} \sqrt{3}\\\\ \text{(d)}\ \ \log _{10} 4 & \text{(e)}\ \ \log _{10} 4.5 & \text{(f)}\ \ \log _{10} 8\\\\ \text{(g)}\ \ \log _{10} 18 & \text{(h)}\ \ \log _{10} 5 \end{array}$

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$\begin{array}{l} {{\log }_{{10}}}2=0.3010,\ {{\log }_{{10}}}3=0.4771\\\\ \text{(a)}\;\;{{\log }_{{10}}}6=\;{{\log }_{{10}}}\left( {2\times 3} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}2+{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+0.4771\\\ \ \ \ \ \ \ \ \ \ \ \ =0.7781\\\\ \text{(b)}\;\;{{\log }_{{10}}}1.5=\;{{\log }_{{10}}}\displaystyle \frac{3}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.4771-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.1761\\\\ \text{(c)}\;\;{{\log }_{{10}}}\sqrt{3}\ =\;{{\log }_{{10}}}{{3}^{{\frac{1}{2}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}\times 0.4771\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.2386\\\\\text{(d)}\;\;{{\log }_{{10}}}4={{\log }_{{10}}}{{2}^{2}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =2\ (0.3010)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6020\\\\ \text{(e)}\;\;{{\log }_{{10}}}4.5=\;{{\log }_{{10}}}\displaystyle \frac{9}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}\displaystyle \frac{{{{3}^{2}}}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2\left( {0.4771} \right)-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.6532\\\\ \text{(f)}\;\;{{\log }_{{10}}}8\ ={{\log }_{{10}}}{{2}^{3}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =3\left( {0.3010} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.9030\\\\ \text{(g)}\;\;{{\log }_{{10}}}18={{\log }_{{10}}}\left( {2\times {{3}^{2}}} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}2+2{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+2\left( {0.4771} \right)\\\\ \text{(h)}\;\;{{\log }_{{10}}}5={{\log }_{{10}}}\displaystyle \frac{{10}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}10-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =1-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6990 \end{array}$

6.           Solve the following equations for $x$.

              $\begin{array}{lll} \text{(a)}\ \ \log _{a} \displaystyle\frac{18}{5}+\log _{a} \displaystyle\frac{10}{3}-\log _{a} \displaystyle\frac{6}{7}=\log _{a} x\\\\ \text{(b)}\ \ \log _{b} x=2-a+\log _{b}\left(\displaystyle\frac{a^{2} b^{a}}{b^{2}}\right)\\\\ \text{(c)}\ \ \log x^{3}-\log x^{2}=\log 5 x-\log 4 x\\\\ \text{(d)}\ \ \log _{10} x+\log _{10} 3=\log _{10} 6\\\\ \text{(e)}\ \ 8 \log x=\log a^{\frac{3}{2}}+\log 2-\displaystyle\frac{1}{2} \log a^{3}-\log \frac{2}{a^{4}}\\\\ \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{a}}\displaystyle \frac{{18}}{5}+{{\log }_{a}}\displaystyle \frac{{10}}{3}-{{\log }_{a}}\displaystyle \frac{6}{7}={{\log }_{a}}x\\ \ \ \ \ {{\log }_{a}}\left( {\displaystyle \frac{{\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}}}{{\displaystyle \frac{6}{7}}}} \right)={{\log }_{a}}x\\ \ \ \ \ x=\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}\times \displaystyle \frac{7}{6}\\ \ \ \ \ x=14\\\\ \text{(b)}\;\;{{\log }_{b}}x=2-a+{{\log }_{b}}\left( {\displaystyle \frac{{{{a}^{2}}{{b}^{a}}}}{{{{b}^{2}}}}} \right)\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+{{\log }_{b}}{{b}^{a}}-{{\log }_{b}}{{b}^{2}}\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+a-2\\ \ \ \ \ {{\log }_{b}}x={{\log }_{b}}{{a}^{2}}\\ \ \ \ \ x={{a}^{2}}\\\\ \text{(c)}\;\;\log {{x}^{3}}-\log {{x}^{2}}=\log 5x-\log 4x\\ \ \ \ \ \log \displaystyle \frac{{{{x}^{3}}}}{{{{x}^{2}}}}=\log \displaystyle \frac{{5x}}{{4x}}\\ \ \ \ \ \log x=\log \displaystyle \frac{5}{4}\\ \ \ \ \ x=\displaystyle \frac{5}{4}\\\\ \text{(d)}\;\;{{\log }_{{10}}}x+{{\log }_{{10}}}3={{\log }_{{10}}}6\\ \ \ \ \ \ {{\log }_{{10}}}3x={{\log }_{{10}}}6\\ \ \ \ \ \ 3x=6\\ \ \ \ \ \ x=2\\ \text{(e)}\;\;8\log x=\log {{a}^{{\frac{3}{2}}}}+\log 2- \frac{1}{2}\log {{a}^{3}}-\log \displaystyle \frac{2}{{{{a}^{4}}}}\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{{\frac{3}{2}}}}+\log 2-\log {{a}^{{ \frac{3}{2}}}}-\left( {\log 2-\log {{a}^{4}}} \right)\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{4}}\\ \ \ \ \ {{x}^{8}}={{a}^{4}}\\ \ \ \ \ x=\sqrt{a}\end{array}$

7.           Given that $\log_{10} 5 = 0.6990$ and $\log_{10}x = 0.2330$. What is the value of $x$?

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$\begin{array}{l} \ \ \;\;{{\log }_{{10}}}5=0.6990\\\\ \ \ \ \ {{\log }_{{10}}}x=0.2330\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x=0.6990\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x={{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x=\displaystyle\frac{1}{3}{{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x={{\log }_{{10}}}{{5}^{{\frac{1}{3}}}}\\\\ \ \ \ \ \therefore \ \ x={{5}^{{\frac{1}{3}}}}=\sqrt[3]{5} \end{array}$

8.           Show that if $\log _{e} I=-\displaystyle\frac{R}{L} t+\log _{e} I_{0}$ then $I=I_{0} e^{-\frac{R t}{L}}$

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$\begin{array}{l} {{\log }_{e}}I=-\displaystyle \frac{R}{L}t+{{\log }_{e}}{{I}_{0}}\\\\ {{\log }_{e}}I-{{\log }_{e}}{{I}_{0}}=-\displaystyle \frac{{Rt}}{L}\\\\ {{\log }_{e}}\displaystyle \frac{I}{{{{I}_{0}}}}=-\displaystyle \frac{{Rt}}{L}\\\\ \displaystyle \frac{I}{{{{I}_{0}}}}={{e}^{{-\frac{{Rt}}{L}}}}\\\\ \therefore \ I={{I}_{0}}{{e}^{{-\frac{{Rt}}{L}}}} \end{array}$

9.           Show that if $\log _{b} y=\displaystyle\frac{1}{2} \log _{b} x+c$ then $y=b^{c} \sqrt{x}$

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$\begin{array}{l}{{\log }_{b}}y=\displaystyle \frac{1}{2}{{\log }_{b}}x+c\\\\ {{\log }_{b}}y-\displaystyle \frac{1}{2}{{\log }_{b}}x=c\\\\ {{\log }_{b}}y-{{\log }_{b}}{{x}^{{\frac{1}{2}}}}=c\\\\ {{\log }_{b}}\displaystyle \frac{y}{{\sqrt{x}}}=c\\\\ \displaystyle \frac{y}{{\sqrt{x}}}={{b}^{c}}\\\\y=\ {{b}^{c}}\sqrt{x} \end{array}$

10.           Show that

              $\begin{array}{l} \text{(a)}\ \ \displaystyle\frac{1}{4} \log _{10} 8+\frac{1}{4} \log _{10} 2=\log _{10} 2\\\\ \text{(b)}\ \ 4 \log _{10} 3-2 \log _{10} 3+1=\log _{10} 90 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ \ \displaystyle \frac{1}{4}{{\log }_{{10}}}8+\displaystyle \frac{1}{4}{{\log }_{{10}}}2\\\ \ \ \ =\displaystyle \frac{1}{4}\left( {{{{\log }}_{{10}}}8+{{{\log }}_{{10}}}2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}\left( {8\times 2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}16\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}{{2}^{4}}\\\ \ \ \ =\displaystyle \frac{1}{4}\times 4{{\log }_{{10}}}2\\\ \ \ \ ={{\log }_{{10}}}2\\\\\text{(b)}\;\;\ \ 4{{\log }_{{10}}}3-2{{\log }_{{10}}}3+1\\\ \ \ \ =2{{\log }_{{10}}}3+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}{{3}^{2}}+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}\left( {{{3}^{2}}\times 10} \right)\\\ \ \ \ ={{\log }_{{10}}}90\end{array}$

11.           Show that

              $\begin{array}{l} \text{(a)}\ \ a^{2 \log _{a} 3}+b^{3 \log _{b} 2}=17\\\\ \text{(b)}\ \ 3 \log _{6} 1296=2 \log _{4} 4096 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ {{a}^{{2{{{\log }}_{a}}3}}}+{{b}^{{3{{{\log }}_{b}}2}}}\\\ \ \ \ ={{a}^{{{{{\log }}_{a}}{{3}^{2}}}}}+{{b}^{{{{{\log }}_{b}}{{2}^{3}}}}}\\\ \ \ \ ={{3}^{2}}+{{2}^{3}}\\\ \ \ \ =17\\\\\text{(b)}\;\ \ 3{{\log }_{6}}1296=3{{\log }_{6}}{{6}^{4}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3\times 4{{\log }_{6}}6\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \ \ \ 2{{\log }_{4}}4096=2{{\log }_{4}}{{4}^{6}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\times 6{{\log }_{4}}4\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \therefore \ \ 3{{\log }_{6}}1296=2{{\log }_{4}}4096\end{array}$

12.           Given that $\log _{10} 12=1.0792$ and $\log _{10} 24=1.3802,$ deduce the values of $\log _{10} 2$ and $\log _{10} 6$.

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$\begin{array}{l}{{\log }_{{10}}}12=1.0792\\\\{{\log }_{{10}}}24=1.3802\\\\{{\log }_{{10}}}2={{\log }_{{10}}}\displaystyle \frac{{24}}{{12}}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}24-{{\log }_{{10}}}12\\\ \ \ \ \ \ \ =1.3802-1.0792\\\ \ \ \ \ \ \ =0.3010\\\\{{\log }_{{10}}}6={{\log }_{{10}}}\displaystyle \frac{{12}}{2}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}12-{{\log }_{{10}}}2\\\ \ \ \ \ \ \ =1.0792-0.3010\\\ \ \ \ \ \ \ =0.7782\end{array}$

13.           If $\log _{x} a=5$ and $\log _{x} 3 a=9$, find the values of $a$ and $x$.

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$\begin{array}{l}{{\log }_{x}}a=5\\\\\therefore \ \ a={{x}^{5}}\\\\{{\log }_{x}}3a=9\\\\\therefore \ \ 3a={{x}^{9}}\\\\\therefore \ \ 3{{x}^{5}}={{x}^{9}}\\\\\therefore \ \ {{x}^{4}}=3\\\\\therefore \ \ x={{3}^{{\frac{1}{4}}}}=\sqrt[4]{3}\\\\\therefore \ \ a={{\left( {{{3}^{{\frac{1}{4}}}}} \right)}^{5}}={{3}^{{\frac{5}{4}}}}=3\sqrt[4]{3}\end{array}$

14.           $\text { (a) }$ If $\log _{10} 2=a,$ find $\log _{10} 8+\log _{10} 25$ in terms of $a$.

   $\text { (b) }$ If $a=10^{x}$ and $b=10^{y},$ express $\log _{10}\left(a^{4} b^{3}\right)$ in terms of $x$ and $y$.

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$\begin{array}{l}(\text{a})\ {{\log }_{{10}}}2=a\ (\text{given})\\\\\ \ \ {{\log }_{{10}}}8+{{\log }_{{10}}}25\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}\frac{{100}}{4}\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}100-{{\log }_{{10}}}4\\\\={{\log }_{{10}}}{{2}^{3}}+{{\log }_{{10}}}{{10}^{2}}-{{\log }_{{10}}}{{2}^{2}}\\\\=3{{\log }_{{10}}}2+2{{\log }_{{10}}}10-2{{\log }_{{10}}}2\\\\={{\log }_{{10}}}2+2\\\\=a+2\\\\(\text{b})\ \ \left. \begin{array}{l}a={{10}^{x}}\\b={{10}^{y}}\end{array} \right\}(\text{given})\\\\\ \ \ \ \ {{\log }_{{10}}}\left( {{{a}^{4}}{{b}^{3}}} \right)={{\log }_{{10}}}\left( {{{{\left( {{{{10}}^{x}}} \right)}}^{4}}{{{\left( {{{{10}}^{y}}} \right)}}^{3}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}\left( {\left( {{{{10}}^{{4x}}}} \right)\left( {{{{10}}^{3}}^{y}} \right)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}{{10}^{{4x+3y}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4x+3y\end{array}$

15.           $\text { (a) }$ If $\log _{2}(4 x-4)=2$, find the value of $\log _{4} x$.

   $\text { (b) }$ Prove that if $\displaystyle\frac{1}{2} \log _{3} M+3 \log _{3} N=1$ then $M N^{6}=9$.

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$\begin{array}{l} (\text{a})\ {{\log }_{2}}(4x-4)=2\\\\ \ \ \ 4x-4={{2}^{2}}\\\\ \ \ \ 4x=8\\\\ \ \ \ x=2\\\\ \ \ \ x={{4}^{{\frac{1}{2}}}}\\\\ \ \ \ {{\log }_{4}}x=\displaystyle\frac{1}{2}\\\\\\ (\text{b})\ \ \displaystyle\frac{1}{2}{{\log }_{3}}M+3{{\log }_{3}}N=1\\\\ \ \ \ \ \ {{\log }_{3}}{{M}^{{\frac{1}{2}}}}+{{\log }_{3}}{{N}^{3}}=1\\\\ \ \ \ \ \ {{\log }_{3}}\left( {{{M}^{{\frac{1}{2}}}}{{N}^{3}}} \right)=1\\\\ \ \ \ \ \ {{M}^{{\frac{1}{2}}}}{{N}^{3}}=3\\\\ \ \ \ \ \ \text{Squaring both sides}\text{.}\\\\ \ \ \ \ \ M{{N}^{6}}=9 \end{array}$