Showing posts with label chapter-3. Show all posts
Showing posts with label chapter-3. 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


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

Tuesday, July 28, 2020

Logarithms : Exercise (3.4) - Solutions

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

Show/Hide Solution

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

2.          Show that

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

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

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

Show/Hide Solution

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

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

Show/Hide Solution

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

Show/Hide Solution

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

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