Slope of a Line : Exercise (1.2) - Solution

Ph မျက်နှာပြင်တွင် စာများအပြည့် မပေါ်လျှင် slider ကို ဆွဲ၍ လည်းကောင်း၊ ph ကို အလျားလိုက်ပုံစံ (landscape position) ပြောင်း၍ လည်းကောင်း ဖတ်ရှုနိုင်ပါသည်။

1.          Complete each sentence.

$\text{(a)}\quad$ The slope of the line passing through two points $(-6, 0)$ and $(2, 3)$ is __________.

$\text{(b)}\quad$ The slope of the line joining the point $(1, 2)$ and the origin is __________.

$\text{(c)}\quad$ A vertical line has __________ slope.

$\text{(d)}\quad$ A horizontal line has __________ slope .

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(a) $m=\displaystyle\frac{3-0}{2-(-6)}=\displaystyle\frac{3}{8}$

(b) $m=\displaystyle\frac{2-0}{1-0}=2$

(c) undefined

(d) zero

2.          For each graph state whether the slope is positive, negative, zero or undefined, then find the slope if possible.

(a)	(b)
(c)	(d)
(e)	(f)
(g)	(h)

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(a) undefined slope

(b) undefined slope

(c) zero slope

(d) zero slope

(e) positive slope, $m=1$

(f) positive slope, $m=\displaystyle\frac{2}{3}$

(g) negative slope, $m=-\displaystyle\frac{3}{4}$

(h) negative slope, $m=-\displaystyle\frac{2}{3}$

3.          Which pairs of points given below will determine horizontal lines? Which ones vertical lines? Determine the slope of each line without calculation.

$\displaystyle \begin{array}{l} \text{(a)}\quad (5,2)\ \text{and}\ (-3,2) \\\\ \text{(b)}\quad (0,5)\ \text{and}\ (-1,5)\\\\ \text{(c)}\quad (2,3)\ \text{and}\ (2,6) \\\\ \text{(d)}\quad (0,0)\ \text{and}\ (0,-2)\\\\ \text{(e)}\quad (1,-2)\ \text{and}\ (-3,-2)\\\\ \text{(f)}\quad (a,b)\ \text{and}\ (a,c) \end{array}$

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(a) horizontal line, slope = 0 ($\because$ same y-coordinate)

(b) horizontal line, slope = 0 ($\because$ same y-coordinate)

(c) vertical line, slope = undefined ($\because$ same x-coordinate)

(d) vertical line, slope = undefined ($\because$ same x-coordinate)

(e) horizontal line, slope = 0 ($\because$ same y-coordinate)

(f) vertical line, slope = undefined ($\because$ same x-coordinate)

4.          Find the slope of each line which contains each pair of points listed below.

$ \displaystyle \begin{array}{l} \text{(a) }\quad A(0,0)\ \text{ and }\ B(8,4)\\\\ \text{(b) }\quad C(10,5)\ \text{ and }\ D(6,8)\\\\ \text{(c) }\quad E(-5,7)\ \text{ and }\ F(-2,-4)\\\\ \text{(d) }\quad G(23,15)\ \text{ and }\ H(18,5)\\\\ \text{(e) }\quad I(-2,0)\ \text{and }\ J(0,\ 6)\\\\ \text{(f) }\quad K(15,6)\ \text{ and }\ L(-2,23) \end{array}$

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The slope of the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $m=\displaystyle\frac{y_2-y_1}{x_2-x_1}.$

$\begin{aligned} \text{(a)}\quad m_{AB}&=\displaystyle\frac{4-0}{8-0}\\\\ &= \displaystyle\frac{1}{2} \end{aligned}$

$\begin{aligned} \text{(b)}\quad m_{CD}&=\displaystyle\frac{8-5}{6-10}\\\\ &= -\displaystyle\frac{3}{4} \end{aligned}$

$\begin{aligned} \text{(c)}\quad m_{EF}&=\displaystyle\frac{-4-7}{-2-(-5)}\\\\ &= -\displaystyle\frac{11}{3} \end{aligned}$

$\begin{aligned} \text{(d)}\quad m_{GH}&=\displaystyle\frac{5-15}{18-23}\\\\ &= \displaystyle\frac{-10}{-5}\\\\ &=2 \end{aligned}$

$\begin{aligned} \text{(f)}\quad m_{IJ}&=\displaystyle\frac{6-0}{0-(-2)}\\\\ &= \displaystyle\frac{6}{2}\\\\ &=3 \end{aligned}$ $\begin{aligned} \text{(f)}\quad m_{KL}&=\displaystyle\frac{23-6}{-2-15}\\\\ &= \displaystyle\frac{17}{-17}\\\\ &=-1 \end{aligned}$

5.          Find the slope of each line which contains each pair of points listed below.

$ \displaystyle \begin{array}{l} \text{(a) }\quad E\left( {\displaystyle\frac{3}{4},\frac{4}{5}\text{ }} \right)\ \text{and }\ F\left( {-\displaystyle\frac{1}{2},\frac{7}{5}} \right)\\\\ \text{(b) }\quad G(-a,b)\ \text{ and }\ H(3a,2b)\\\\ \text{(c) }\quad L\left( {\sqrt{{12}},\sqrt{{18}}} \right)\ \text{ and }\ M\left( {\sqrt{{27}},\sqrt{8}} \right)\\\\ \text{(d) }\quad P(0,a)\ \text{ and }\ Q(a,0) \end{array}$

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The slope of the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $m=\displaystyle\frac{y_2-y_1}{x_2-x_1}.$

$\begin{aligned} \text{(a)}\quad m_{EF}&=\displaystyle\frac{\frac{7}{5}-\frac{4}{5}}{-\frac{1}{2}-\frac{3}{4}}\\\\ &= \displaystyle\frac{\frac{3}{5}}{-\frac{5}{4}}\\\\ &= -\displaystyle\frac{12}{25} \end{aligned}$

$\begin{aligned} \text{(b)}\quad m_{CD}&=\displaystyle\frac{2b-b}{3a-(-a)}\\\\ &= \displaystyle\frac{b}{4a} \end{aligned}$

$\begin{aligned} \text{(c)}\quad m_{EF}&=\displaystyle\frac{\sqrt{8}-\sqrt{18}}{\sqrt{27}-\sqrt{12}}\\\\ &= \displaystyle\frac{2\sqrt{2}-3\sqrt{2}}{3\sqrt{3}-2\sqrt{3}}\\\\ &= \displaystyle\frac{-\sqrt{2}}{\sqrt{3}}\\\\ &= -\displaystyle\frac{\sqrt{6}}{3} \end{aligned}$

$\begin{aligned} \text{(d)}\quad m_{GH}&=\displaystyle\frac{0-a}{a-0}\\\\ &= \displaystyle\frac{-a}{a}\\\\ &=-1 \end{aligned}$

6.          Find $p, q, r$ in the followings:

$\text{(a) }\quad$ The slope joining the points $(0,3)$ and $(1,p)$ is $5$.

$\text{(b) }\quad$ The slope joining the points $(-2, q)$ and $(0,1)$ is $-1$.

$\text{(c) }\quad$ The slope joining the points $(-4, -2)$ and $(r, -6)$ is $-6$.

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The slope of the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $m=\displaystyle\frac{y_2-y_1}{x_2-x_1}.$

$\begin{aligned} \text{(a)}\quad \displaystyle\frac{p-3}{1-0}&=5\\\\ p-3 &= 5\\\\ p &= 8 \end{aligned}$

$\begin{aligned} \text{(b)}\quad \displaystyle\frac{1-q}{0-(-2)}&=-1\\\\ \displaystyle\frac{1-q}{2}&=-1\\\\ 1-q&=-2\\\\ q & = 3 \end{aligned}$

$\begin{aligned} \text{(c)}\quad \displaystyle\frac{-6-(-2)}{r-(-4)}&=-6\\\\ \displaystyle\frac{-4}{r+4}&=-6\\\\ r+4&=\displaystyle\frac{-4}{-6}\\\\ r+4&=\displaystyle\frac{2}{3}\\\\ r&=-\displaystyle\frac{10}{3} \end{aligned}$

7.          Find the slope corresponding to the following events.

$\text{(a) }\quad$ A man climbs $10$ m for every $200$ meters horizontally.

$\text{(b) }\quad$ A motorbike rises $20$ km for every $100$ kilometers horizontally.

$\text{(c) }\quad$ A plane takes off $35$ km for every $5$ kilometers horizontally.

$\text{(d) }\quad$ A submarine descends $120$ m for every $15$ meters horizontally.

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(a) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{10}{200}=\displaystyle\frac{1}{20} $

(b) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{20}{100}=\displaystyle\frac{1}{5} $

(c) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{35}{5}=7 $

(d) $m=\displaystyle\frac{\text{rise}}{\text{run}} =\displaystyle\frac{-120}{15}=-8 $

8.          A train climbs a hill with slope $0.05$. How far horizontally has the train travelled after rising $15$ meters?

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$m = 0.05$, rise $= 15 m$
Since $m=\displaystyle\frac{\text{ rise}}{\text{ run}}$,
$0.05=\displaystyle\frac{15}{\text{ run}} \Rightarrow\text{ run} = 300.$

9.          The vertices of a triangle are the points $A(-2,3)$, $B(5,-4)$ and $C(1,8)$. Find the slope of each side and perimeter of a triangle.

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$A=(-2,3),\ B=(5,-4),\ C=(1,8)$

Since $m=\displaystyle\frac{y_2-y_1}{x_2-x_1},$

$m_{AB}=\displaystyle\frac{-4-3}{5+2}=-1$

$m_{BC}=\displaystyle\frac{8+4}{1-5}= -3$

$m_{AC}=\displaystyle\frac{8-3}{1+2}=\frac{5}{3}$

length of a segment $=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\therefore\ AB=\sqrt{(5+2)^2+(-4-3)^2}=\sqrt{98}=9.9$

$BC=\sqrt{(1-5)^2+(8+4)^2}=\sqrt{160}=12.6$

$AC=\sqrt{(1+2)^2+(8-3)^2}=\sqrt{34}=5.8$

$\therefore\ \text{ the perimeter of}\ \triangle ABC = AB + BC + AC = 9.9+12.7+5.8=28.4$

10.          The vertices of a parallelogram are the points $P(1,4)$, $Q(3,2)$, $R(4,6)$ and $S(2,8)$. Find the slope of each side.

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$P=(1,4)$, $Q=(3,2)$, $R=(4,6), S=(2,8)$

Since $m=\displaystyle\frac{y_2-y_1}{x_2-x_1},$

$m_{PQ}=\displaystyle\frac{2-4}{3-1}=-1$

$m_{QR}=\displaystyle\frac{6-2}{4-3}= 4$

$m_{RS}=\displaystyle\frac{8-6}{2-4}=-1$

$m_{PS}=\displaystyle\frac{8-4}{2-1}=4$

11.          A line having a slope of $-1$ contains the point $(-2,5)$. What is the $y$-coordinate of the point on that line whose $x$-coordinate is $8$?

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Let the required point be $(8, y).$

Since $m=\displaystyle\frac{y_2-y_1}{x_2-x_1},$

$-1=\displaystyle\frac{y-5}{8+2}$

$\therefore\ y= -5$.

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) $\log 0.0002345$

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

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

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

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

(a) $\log_{8}25$

(b) $\log_{5}64$

(c) $\log_{25}100$

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

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

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 $a b c=1$.

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

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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}$

Exponents and Radicals : Exercise (2.3) - Solutions

1.         Write the following in radical form.

$\begin{array}{ll} \text{(a)}\quad (5)^{^{\tfrac{1}{2}}} & \text{(b)}\quad (-9)^{^{\tfrac{1}{3}}} \\\\ \text{(c)}\quad (2)^{^{-\frac{1}{2}}} & \text{(d)}\quad \left(-\displaystyle\frac{3}{4}\right)^{^{\tfrac{2}{5}}}\\\\ \text{(e)}\quad \left(\displaystyle\frac{2}{7}\right)^{^{\tfrac{5}{2}}} \end{array}$

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$\begin{array}{l} \text{(a)}\ \ {{\left( 5 \right)}^{{\tfrac{{1\,}}{{2\,}}}}}=\sqrt{5}\\\\ \text{(b)}\ \ {{\left( {-9} \right)}^{{\tfrac{1}{3}}}}=\sqrt[3]{{-9}}\\\\ \text{(c)}\ \ {{\left( 2 \right)}^{{-\tfrac{{1\,}}{{2\,}}}}}=\displaystyle \frac{1}{{{{{\left( 2 \right)}}^{{\tfrac{{1\,}}{{2\,}}}}}}}=\displaystyle \frac{1}{{\sqrt{2}}}\\\\ \text{(d)}\ \ {{\left( {-\displaystyle \frac{3}{4}} \right)}^{{\tfrac{2}{5}}}}=\sqrt[5]{{{{{\left( {-\displaystyle \frac{3}{4}} \right)}}^{2}}}}=\sqrt[5]{{\displaystyle \frac{9}{{16}}}}\\\\ \text{(e)}\ \ {{\left( {\displaystyle \frac{2}{7}} \right)}^{{\tfrac{5}{2}}}}=\sqrt{{{{{\left( {\displaystyle \frac{2}{7}} \right)}}^{5}}}}=\sqrt{{\displaystyle \frac{{32}}{{16807}}}}\\ \end{array}$

2.         Write the following in fractional exponent form.

$\begin{array}{ll} \text{(a)}\quad \sqrt[6]{c^{^{5}}} & \text{(b)}\quad \sqrt[3]{-2} \\\\ \text{(c)}\quad \sqrt[5]{a^{^{4}} \sqrt[3]{b^{^{5}}}} & \text{(d)}\quad \sqrt[4]{\left(\displaystyle\frac{3}{7}\right)^{^{3}}} \end{array}$

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$\begin{array}{l} (\text{a})\ \ \sqrt[6]{{{{c}^{5}}}}={{c}^{{\frac{5}{6}}}}\\\\ (\text{b})\ \ \sqrt[3]{{-2}}={{(-2)}^{{\frac{1}{3}}}}\\\\ (\text{c})\ \ \sqrt[5]{{{{a}^{4}}\sqrt[3]{{{{b}^{5}}}}}}=\sqrt[5]{{{{a}^{4}}{{b}^{{\frac{5}{3}}}}}}={{\left( {{{a}^{4}}{{b}^{{\frac{5}{3}}}}} \right)}^{{\frac{1}{5}}}}={{a}^{{\frac{4}{5}}}}{{b}^{{\frac{1}{3}}}}\\\\ (\text{d})\ \ \sqrt[4]{{{{{\left( {\displaystyle \frac{3}{7}} \right)}}^{3}}}}={{\left( {\displaystyle \frac{3}{7}} \right)}^{{\frac{3}{4}}}} \end{array}$

3.         Change the expression with the same radical and simplify the radicands.

$\begin{array}{ll} \text{(a)}\quad 6 \sqrt{2} \quad& \text{(b)}\quad 3 a \sqrt[3]{x} \\\\ \text{(c)}\quad 2 \sqrt[5]{2} \quad& \text{(d)}\quad \sqrt[4]{\displaystyle\frac{1}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{^{3}}} \end{array}$

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$\begin{array}{l} \text{(a)}\quad 6 \sqrt{2}=\sqrt{6^{2} \cdot 2}=\sqrt{72}\\\\ \text{(b)}\quad 3 a \sqrt[3]{x}=\sqrt[3]{3^{3} \cdot a^{3} x}=\sqrt[3]{27 a^{3} x}\\\\ \text{(c)}\quad 2 \sqrt[5]{2}=\sqrt[5]{2^{5} \cdot 2}=\sqrt[5]{64}\\\\ \text{(d)}\quad 3 \sqrt[4]{\displaystyle\frac{1}{2}}=\sqrt[4]{3^{4} \cdot \displaystyle\frac{1}{2}}=\sqrt[4]{\displaystyle\frac{81}{2}}\\\\ \text{(e)}\quad 3 \sqrt{x^{3}}=\sqrt{3^{2} \cdot x^{3}}=\sqrt{9 x^{3}} \end{array}$

4.         Simplify.

$\begin{array}{ll} \text{(a)}\quad \sqrt{32} & \text{(b)}\quad \sqrt[5]{-32} \\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{^{16}}}{16 y^{^{4}}}} & \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{^{2}}}{4 y}} \\\\ \text{(e)}\ \displaystyle\frac{9^{^{\tfrac{1}{2}}}}{\sqrt[3]{27}}& \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{^{3}}}} & \text{(h)}\quad \sqrt[n]{\displaystyle\frac{32}{2^{^{5+n}}}} \end{array}$

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$\begin{array}{l} \text{(a)}\quad \sqrt{32}=\sqrt{16 \cdot 2}=4 \sqrt{2}\\\\ \text{(b)}\quad \sqrt[5]{-32}=\sqrt[5]{(-2)^{5}}=-2\\\\ \text{(c)}\quad \sqrt[4]{\displaystyle\frac{81 x^{16}}{16 y^{4}}}=\sqrt[4]{\displaystyle\frac{3^{4}\left(x^{4}\right)^{4}}{2^{4} \cdot y^{4}}}=\displaystyle\frac{3 x^{4}}{2 y}\\\\ \text{(d)}\quad \sqrt[3]{\displaystyle\frac{81 x^{2}}{4 y}}=\sqrt[3]{\displaystyle\frac{3^{3} \cdot 3 x^{2}}{4 y}}=3 \sqrt[3]{\displaystyle\frac{3 x^{2}}{4 y}}\\\\ \text{(e)}\quad \displaystyle\frac{9^{\displaystyle\frac{1}{2}}}{\sqrt[3]{27}}=\displaystyle\frac{\left(3^{2}\right)^{\frac{1}{2}}}{\sqrt[3]{3^{3}}}=\displaystyle\frac{3}{3}=1\\\\ \text{(f)}\quad \sqrt{\displaystyle\frac{2}{3}} \cdot \sqrt{\displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{2}{3} \times \displaystyle\frac{75}{98}}=\sqrt{\displaystyle\frac{25}{49}}=\displaystyle\frac{5}{7}\\\\ \text{(g)}\quad \sqrt[3]{\displaystyle\frac{-216}{8 \times 10^{3}}}=\sqrt[3]{\displaystyle\frac{(-6)^{3}}{2^{3} \times 10^{3}}}=\displaystyle\frac{-6}{2 \times 10}=-\displaystyle\frac{3}{10}\\\\ \text{(h)}\quad \sqrt[4]{\displaystyle\frac{32}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{2^{5}}{2^{5+n}}}=\sqrt[n]{\displaystyle\frac{1}{2^{n}}}=\displaystyle\frac{1}{2} \end{array}$

5.         Rationalize the denominators.

$\begin{array}{ll} \text{(a)}\quad \displaystyle\frac{4 \sqrt{35}}{3 \sqrt{7}}\quad & \text{(b)}\quad \displaystyle\frac{20}{\sqrt{5}} \\\\ \text{(c)}\quad \displaystyle\frac{18}{\sqrt[3]{2}}\quad& \text{(d)}\quad \displaystyle\frac{\sqrt[3]{32}}{\sqrt[4]{27}} \\\\ \text{(e)}\quad \displaystyle\frac{\sqrt[3]{36 a^{^{2}}}}{\sqrt[3]{9 a}}\quad & \text{(f)}\quad \displaystyle\frac{\sqrt[3]{2}}{\sqrt[6]{12}} \\\\ \text{(g)}\quad \displaystyle\frac{1}{\sqrt[3]{x y^{^{2}}}}\quad & \text{(h)}\quad \sqrt[m]{{\displaystyle\frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}} \end{array}$

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$\begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \frac{{4\sqrt{{35}}}}{{3\sqrt{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{35}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{4}{3}\sqrt{{\displaystyle \frac{{7\times 5}}{7}}}\\\\\ \ \ \ \ =\displaystyle \frac{{4\sqrt{5}}}{3}\\\\\\\text{(b)}\ \ \ \ \displaystyle \frac{{20}}{{\sqrt{5}}}\\\\\ \ \ \ \ =\displaystyle \frac{{20}}{{\sqrt{5}}}\times \displaystyle \frac{{\sqrt{5}}}{{\sqrt{5}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{20\sqrt{5}}}{5}\\\\\ \ \ \ \ =4\sqrt{5}\\\\\\\text{(c)}\ \ \ \ \displaystyle \frac{{18}}{{\sqrt[3]{2}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18}}{{\sqrt[3]{2}}}\times \displaystyle \frac{{\sqrt[3]{{{{2}^{2}}}}}}{{\sqrt[3]{{{{2}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{{\sqrt[3]{{{{2}^{3}}}}}}\ \\\\\ \ \ \ \ =\displaystyle \frac{{18\sqrt[3]{4}}}{2}\\\ \ \ \ \ =9\sqrt[3]{4}\\\\\text{(d)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{32}}}}{{\sqrt[4]{{27}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{2}^{3}}\cdot 4}}}}{{\sqrt[4]{{{{3}^{3}}}}}}\times \displaystyle \frac{{\sqrt[4]{3}}}{{\sqrt[4]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{2\sqrt[3]{4}\sqrt[4]{3}}}{3}\\\\\\\text{(e)}\ \ \ \ \displaystyle \frac{{\sqrt[3]{{36{{a}^{2}}}}}}{{\sqrt[3]{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{\displaystyle \frac{{36{{a}^{2}}}}{{9a}}}}\\\\\ \ \ \ \ =\sqrt[3]{{4a}}\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{12}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[6]{{{{2}^{2}}}}\cdot \sqrt[6]{3}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{2}}}{{\sqrt[3]{2}\cdot \sqrt[6]{3}}}\times \displaystyle \frac{{\sqrt[6]{{{{3}^{5}}}}}}{{\sqrt[6]{{{{3}^{5}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[6]{{243}}}}{3}\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{1}{{\sqrt[3]{{x{{y}^{2}}}}}}\times \displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{2}}y}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{\sqrt[3]{{{{x}^{3}}{{y}^{3}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[3]{{{{x}^{2}}y}}}}{{xy}}\\\\\text{(h)}\ \ \ \ \ \sqrt[m]{{\displaystyle \frac{{2{{x}^{2}}{{y}^{{3m}}}}}{{9{{x}^{5}}{{y}^{{4m-1}}}}}}}\\\\\ \ \ \ \ =\sqrt[m]{{\displaystyle \frac{{2y}}{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\ \displaystyle \frac{{\sqrt[m]{{2y}}}}{{\sqrt[m]{{{{3}^{2}}{{x}^{3}}{{y}^{m}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2y}}}}{{y\sqrt[m]{{{{3}^{2}}{{x}^{3}}}}}}\times \displaystyle \frac{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}{{\sqrt[m]{{{{3}^{{m-2}}}{{x}^{{m-3}}}}}}}\\\\\ \ \ \ \ =\displaystyle \frac{{\sqrt[m]{{2\cdot {{3}^{{m-2}}}{{x}^{{m-3}}}y}}}}{{3xy}}\end{array}$

6.         Reduce the order as far as possible.

$\begin{array}{ll} \text{(a)}\quad \sqrt[4]{25} \quad& \text{(b)}\quad \sqrt[6]{4} \\\\ \text{(c)}\quad \sqrt[6]{8} \quad& \text{(d)}\quad \sqrt[9]{8 y^{^{3}}} \\\\ \text{(e)}\quad \sqrt[6]{27^{^{3}}} \quad& \text{(f)}\quad \sqrt[8]{a^{2} b^{^{4}}} \\\\ \text{(g)}\quad \sqrt[12]{64 a^{^{2}} b^{^{6}}} & \text{(h)}\quad (72)^{^{\tfrac{3}{5}}} \\\\ \text{(i)}\quad \sqrt[3]{768} \end{array}$

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$\begin{array}{l} \text{(a)}\quad \sqrt[4]{25}=\sqrt[4]{5^{2}}=\sqrt{5}\\\\ \text{(b)}\quad \sqrt[6]{4}=\sqrt[6]{2^{2}}=\sqrt[3]{2}\\\\ \text{(c)}\quad \sqrt[6]{8}=\sqrt[6]{2^{3}}=\sqrt{2}\\\\ \text{(d)}\quad \sqrt[9]{8 y^{3}}=\sqrt[9]{2^{3} \cdot y^{3}}=\sqrt[3]{2 y}\\\\ \text{(e)}\quad \sqrt[6]{27^{3}}=\sqrt{27}=3 \sqrt{3}\\\\ \text{(f)}\quad \sqrt[8]{a^{2} b^{4}}=\sqrt[4]{a b^{2}}\\\\ \text{(g)}\quad \sqrt[12]{64 a^{2} b^{6}}=\sqrt[12]{8^{2} \cdot a^{2} b^{6}}=\sqrt[6]{8 a b^{3}}\\\\ \text{(h)}\quad (72)^{\frac{3}{5}}=\sqrt[5]{72^{3}}=\sqrt[5]{\left(3^{2} \cdot 2^{3}\right)^{3}}=\sqrt[5]{3^{6} \cdot 2^{9}}=6 \sqrt[5]{3 \cdot 16}=6 \sqrt[5]{48}\\\\ \text{(i)}\quad \sqrt[3]{768}=\sqrt[3]{4^{3} \cdot 12}=4 \sqrt[3]{12} \end{array}$

7.         Find the simplified forms.

$\begin{array}{ll} \text{(a)}\quad \sqrt{\displaystyle\frac{9}{50}} \quad& \text{(b)}\quad \sqrt[3]{\displaystyle\frac{-192}{49}} \\\\ \text{(c)}\quad \sqrt[4]{16}\quad& \text{(d)}\quad 2 \sqrt[3]{56} \end{array}$

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$\begin{array}{l} \text{(a)}\quad\sqrt{\displaystyle \frac{9}{50}}=\sqrt{\displaystyle \frac{3^{2}}{5^{2} \cdot 2}}=\displaystyle \frac{3}{5 \sqrt{2}} \times \displaystyle \frac{\sqrt{2}}{\sqrt{2}}=\displaystyle \frac{3 \sqrt{2}}{10}\\\\ \text{(b)}\quad\sqrt[3]{\displaystyle \frac{-192}{49}}=\sqrt[3]{\displaystyle \frac{(-4)^{3} \cdot 3}{7^{2}} \times \displaystyle \frac{7}{7}}=-\displaystyle \frac{4 \sqrt[3]{21}}{7}\\\\ \text{(c)}\quad\sqrt[4]{16}=\sqrt[4]{2^{4}}=2\\\\ \text{(d)}\quad 2 \sqrt[3]{56}=2 \sqrt[3]{2^{3} \cdot 7}=4 \sqrt[3]{7} \end{array}$

Similarity : Exercise (8.1) - Solutions

1.           State why the two polygons are, are not, similar.

(a)
The two rectangles are not similar because corresponding sides are not proportional.

(b)
The two squares are similar because corresponding sides are proportional and corresponding angles are equal.

(c)
The two triangles are similar because corresponding sides are proportional and corresponding angles are equal.

(d)
The two polygons are not similar because they have different number of sides.

2.           Complete the proportions.

             (a)    If $\triangle A B C \sim \triangle D E F$ then $\displaystyle\frac{A B}{?}=\frac{B C}{?}=\frac{?}{D F}$.

             (b)     If $\triangle G H I \sim \triangle K L M$ then $\displaystyle\frac{?}{H I}=\frac{?}{G H}=\frac{?}{G I}$.

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(a) $\displaystyle\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$,

(b) $\displaystyle\frac{LM}{HI}=\frac{KL}{GH}=\frac{KM}{GI}$

3.           State whether the proportions are correct for the indicated similar triangles.

           $\begin{array}{l} \text{(a)}\ \triangle A B C \sim \triangle X Y Z.\\\\ \quad\ \ \displaystyle\frac{A B}{X Y}=\displaystyle\frac{B C}{Y Z}\\\\ \text{(b)}\ \triangle D E F \sim \triangle H I J.\\\\ \quad\ \ \displaystyle\frac{D E}{H I}=\displaystyle\frac{E F}{IJ}\\\\ \text{(c)}\ \triangle R S T \sim \triangle L M K.\\\\ \quad\ \ \displaystyle\frac{R T}{L M}=\displaystyle\frac{S T}{M K}\\\\ \text{(d)}\ \triangle X Y Z \sim \triangle U VW.\\\\ \quad\ \ \displaystyle\frac{X Y}{U V}=\displaystyle\frac{X Z}{V W} \end{array}$

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(a) correct

(b) correct

(c) incorrect

(d) correct

4.           Given    :     $\triangle P Q R \sim \triangle U V W$ and lengths of sides are as marked.

Find      :     The values of $x$ and $y$.

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$\begin{array}{l} \triangle P Q R\sim\triangle U V W\ \ \text { (Given) } \\\\ \displaystyle \frac{P Q}{U V}=\displaystyle \frac{Q R}{V W}=\displaystyle \frac{P R}{U W} \\\\ \displaystyle \frac{7}{x}=\displaystyle \frac{10}{y}=\displaystyle \frac{12}{9} \\\\ \displaystyle \frac{7}{x}=\displaystyle \frac{4}{3} \\\\ 4 x=21 \\\\ x=5.25 \\\\ \displaystyle \frac{10}{y}=\displaystyle \frac{4}{3} \\\\ 4 y=30 \\\\ y=7.5 \end{array}$

5.           The measures of two angles of $\triangle X Y Z$ are $82^{\circ}$ and $16^{\circ} .$ Find the measures of the angles of a triangle similar to $\triangle X Y Z$.

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The measures of the first two angles of $\triangle X Y Z$ are $82^{\circ}$ and $16^{\circ}$.

$\therefore$ The measure of the third angle of $\triangle X Y Z$ \[ \begin{array}{l} =180^{\circ}-\left(82^{\circ}+16^{\circ}\right)\\\\ =180^{\circ}-98^{\circ} \\\\ =82^{\circ} \end{array} \] $\therefore$ The measures of the angles of a triangle similar to $\triangle X Y Z$ are $82^{\circ}, 82^{\circ}$ and $16^{\circ}$ .

Target Mathematics (သင်ရိုးသစ် ဆယ်တန်း သင်္ချာ) - အတွဲ (၁) နှင့် အတွဲ (၂)

မြန်မာနိုင်ငံ၏ ပညာရေးစနစ်ကို K+12 စနစ်သစ်သို့ အတန်းလိုက် နှစ်အပိုင်းအခြားအလိုက် ပြောင်းလဲ လာခဲ့ရာ ၂၀၂၀-၂၁ ပညာသင်နှစ်တွင် အထက်တန်းဆင့် ပြောင်းလဲမှုမှာ အသက်ဝင်လာပြီ ဖြစ်သည်။

အတန်းစနစ် ပြောင်းလဲမှုနှင့်အတူ သင်ခန်းစာသင်ရိုးများလဲ ပြောင်းလဲလာခဲ့ရာ Grade 10 Mathematics လည်းပါဝင်လာပါသည်။ Grade 10 Mathematics သည် ယခင် န၀မတန်း သင်ရိုးကို အခြေခံ ပြောင်းလဲထားပြီး နိုင်ငံတကာ သင်ရိုးများနှင့် လိုက်လျောညီထွေ ဖြစ်စေရန် ပြင်ဆင်ပြောင်းလဲမှု အများအပြား ပါဝင်လာပါသည်။

သင်ရိုးအပြောင်းအလဲတွင် သင်ယူလေ့လာသူ ကျောင်းသား၊ ကျောင်းသူများ သင်ခန်းစာဆိုင်ရာ သဘောတရားများ၊ ဝေါဟာရများ၊ လွယ်ကူစွာ လေ့လာနိုင်ရေးအတွက် Target Mathematics မှ သင်ရိုးသစ် သင်္ချာအထောက်ကူပြု စာအုပ်ကို စီစဉ် ထုတ်ဝေလိုက်ပါသည်။ သင်ရိုးသစ် သင်္ချာတွင် ယခင်သင်ရိုးဟောင်း ကဲ့သို့ အခန်း (၁၀) ခန်းပါဝင်ပါသည်။

သင်ရိုးသစ်ပြဌာန်းချက်တွင် ပါဝင်သော သင်ခန်းစာများ

Chapter	Title
Chapter (1)	Introduction to Coordinate Geometry
Chapter (2)	Exponents and Radicals
Chapter (3)	Logarithms
Chapter (4)	Functions
Chapter (5)	Quadratic Functions
Chapter (6)	Absolute Value Functions
Chapter (7)	Probability
(chapter (8)	Similarity
Chapter (9)	Circles
Chapter (10)	Trigonometry

Target Mathematics သည် အထက်ပါ သင်ခန်းစားများမှ Chapter (1) မှ Chapter (6) အထိကို အတွဲ (၁) ဖြစ် ထုတ်ဝေခဲ့ပြီး Chapter (7) မှ (10) အထိကို အတွဲ(၂) အဖြစ်ဆက်လက် ထုတ်ဝေလိုက်ပါပြီ။ အတွဲ(၁) နှင့် အတွဲ(၂) နှစ်အုပ်လုံးအတွက် အောက်ပါအတိုင်း စီစဉ်ထားပါသည်။

Grade 10 - Vol (1), Price = K 3500	Grade 10 - Vol (2), Price = K 3000

• သင်ခန်းစာ ရှင်းလင်းချက်

• သင်ရိုးအတိုင်း လေ့ကျင့်ခန်းများ

• ထပ်ဆောင်းလေ့ကျင့်ခန်းများ

• သင်ရိုးအကျဉ်းချုပ် စုစည်းမှု (Summary)

• လေ့ကျင့်ခန်းတိုင်း အတွက် answer key များ ဖြစ်ပါသည်။

Target Mathematics အတွဲ (၁) နှင့် အတွဲ(၂) နှစ်အုပ်လုံးသည် လေ့လာသူအတွက် ...

• ပြဌာန်းသင်ရိုးစာအုပ် အဖြစ်သော်လည်းကောင်း

• အထူးထုတ်အသွင် အဖြစ်သော်လည်းကောင်း

• ဖတ်မှတ်ရန် ဂျာနယ် အဖြစ်သော်လည်းကောင်း

• အဖြေတိုက်စစ်ဆေးရန် Key Book အဖြစ်သော်လည်းကောင်း

ဘက်စုံအသုံးချနိုင်ရန် ရေးသားပြင်ဆင် ပေးထားသဖြင့် သင်ယူလေ့လာသူ ကျောင်းသား၊ ကျောင်းသူတို့အတွက် များစွာအထောက်အကူ ဖြစ်စေလိမ့်မည် ဟုယုံကြည်မိပါသည်။

ထို့အပြင် သင်ခန်းစာ လေ့ကျင့်ခန်းတိုင်းအတွက် အဖြေအပြည့်အစုံ (Complete Solution) ကိုလည်း ယခု website တွင် တစ်နှစ်ပတ်လုံး စဉ်ဆက်မပြတ် တင်ပေးသွားမည်ဖြစ်ကြောင်း သတင်းကောင်း ပါးအပ်ပါသည်။

မှာယူလိုပါက အောက်ပါ Order Form တွင် ပုံစံဖြည့်၍ မှာယူနိုင်ပါသည်။ ပုံစံဖြည့်၍ Submit လုပ်လိုက်သည်နှင့် မှာယူသူ၏ အမှာစာကို လက်ခံရပြီး ငွေလွှဲရန်အတွက် ပြန်လည်ဆက်သွယ်ပေးမည် ဖြစ်ပါသည်။