Question 14 Marks
If $m = \log 20$ and $n = \log 25,$ find the value of $x,$ so that $:2 \log (x - 4) = 2\ m - n.$
AnswerGiven that
$\mathrm{m}=\log 20$ and $\mathrm{n}=\log 25$
We also have
$2 \log (x-4)=2 m-n$
$ \Rightarrow 2 \log (x-4)=2 \log 20-\log 25$
$ \Rightarrow \log (x-4)^2=\log 20^2-\log 25$
$ \Rightarrow \log (x-4)^2=\log 400-\log 25$
$ \Rightarrow \log (x-4)^2=\log \frac{400}{25}$
$ \Rightarrow(x-4)^2=\frac{400}{25}$
$ \Rightarrow(x-4)^2=16$
$ \Rightarrow x-4=4$
$ \Rightarrow x=4+4$
$ \Rightarrow x=8 .$
View full question & answer→Question 24 Marks
If $a^2+b^2=23 a b$, show that :$\log \frac{a+b}{5}=\frac{1}{2}(\log a+\log b)$
AnswerGiven that
$a^2+b^2=23 a b$
$ \Rightarrow a^2+b^2+2 a b=23 a b+2 a b$
$ \Rightarrow a^2+b^2+2 a b=25 a b$
$ \Rightarrow(a+b)^2=25 a b$
$ \Rightarrow \frac{(a+b)^2}{25}=a b$
$ \Rightarrow\left(\frac{a+b}{5}\right)^2=a b$
$ \Rightarrow \log \left(\frac{a+b}{5}\right)^2=\mathrm{ab}$
$ \Rightarrow 2 \log \left(\frac{a+b}{5}\right)=\log \mathrm{ab}$
$ \Rightarrow \log \left(\frac{a+b}{5}\right)=\log \mathrm{ab}$
$ \Rightarrow \log \left(\frac{a+b}{5}\right)=\frac{1}{2}(\log a+\log b)$
View full question & answer→Question 34 Marks
If $x=\log 0.6 ; y=\log 1.25$ and $z=\log 3-2 \log 2$, find the values of:$(i) x+y- z,(ii) 5^{x + y - z}$
AnswerGiven that
$x=\log 0.6, y=\log 1.25, z=\log 3-2 \log 2$
Consider
$ z=\log 3-2 \log 2$
$ =\log 3-\log 2^2$
$ =\log 3-\log 4$
$ =\log \frac{3}{4}$
$=\log 0.75\ldots(1)$
$ \text { (i) } x+y-z=\log 0.6+\log 1.25-\log 0.75$
$ =\log \frac{0.6 \times 1.25}{0.75}$
$ =\log \left[\frac{0.75}{0.75}\right]$
$ =\log 1$
$=0\ldots(2)$
$(ii) 5^{x+y-z}=5^0 \ldots[\because x+y-z=0$ from $(2) ]$
$=1$
View full question & answer→Question 44 Marks
Evaluate $: \frac{1}{\log _a b c+1}+\frac{1}{\log _b c a+1}+\frac{1}{\log _c a b+1}$
Answer$ \Rightarrow \frac{1}{\log _a b c+1}+\frac{1}{\log _b c a+1}+\frac{1}{\log _c a b+1}$
$ \Rightarrow \frac{1}{\log _a b c+\log _a a}+\frac{1}{\log _b c a+\log _b b}+\frac{1}{\log _c a b+\log _c c}$
$ \Rightarrow \frac{1}{\log _a a b c}+\frac{1}{\log _b a b c}+\frac{1}{\log _c a b c} \ldots\left[\because \log _{\mathrm{a}} \mathrm{b}+\log _{\mathrm{a}} \mathrm{c}=\log _{\mathrm{a}} \mathrm{bc}\right]$
$ \Rightarrow \frac{1}{\frac{\log a b c}{\log a}}+\frac{1}{\frac{\log a b c}{\log b}}+\frac{1}{\frac{\log a b c}{\log c}}$
$ \Rightarrow \frac{\log a+\log b+\log c}{\log a b c}$
$ \Rightarrow \frac{\log a b c}{\log a b c} \ldots \left[\because \log _{\mathrm{a}} \mathrm{b}+\log _{\mathrm{a}} \mathrm{c}=\log _{\mathrm{a}} \mathrm{bc}\right]$
$ \Rightarrow 1$
View full question & answer→Question 54 Marks
Given : $\frac{\log x}{\log y}=\frac{3}{2}$ and $\log (x y)=5$; find the value of $x$ and $y$.
Answer$ \frac{\log x}{\log y}=\frac{3}{2}$
$ \Rightarrow 2 \log x=3 \log y$
$\Rightarrow \log y=\frac{2 \log x}{3}\ldots(1)$
$ \log (x y)=5$
$ \Rightarrow \log x+\log y=5$
$\Rightarrow \log x+\frac{2 \log x}{3}=5 \ldots .[$ Substituting $(1) ]$
$\Rightarrow \frac{3 \log x+2 \log x}{3}=5$
$\Rightarrow \frac{5 \log x}{3}=5$
$ \Rightarrow \log x=3$
$ \Rightarrow x=10^3$
$ \therefore x=1000$
Substituting $x=1000$
$\log y=\frac{2 \times 3}{3}$
$ \Rightarrow \log y=2$
$ \Rightarrow y=10^2$
$ \therefore y=100 .$
View full question & answer→Question 64 Marks
Prove that : $\log_{10} 125 = 3(1 - \log_{10}2).$
Answer$ \log _{10} 125=3\left(1-\log _{10} 2\right)$
$\text { L.H.S. }=\log _{10} 125$
$\Rightarrow \log _{10} 5 \times 5 \times 5$
$\Rightarrow \log _{10} 5^3$
$\Rightarrow 3 \log _{10} 5 \ldots .(1)$
$\text { R.H.S }=3\left(1-\log _{10} 2\right)$
$\left.\Rightarrow 3 \log _{10} 10-\log _{10} 2\right)$
$\Rightarrow 3 \log _{10}\left(\frac{10}{2}\right)$
$\Rightarrow 3 \log _{10} 5$
From $(1)$ and $(2),$ we have
$\text{L.H.S.} =\text {R. H. S}.$
Hence proved.
View full question & answer→Question 74 Marks
Given $\log _x 25-\log _x 5=2-\log _x\left(\frac{1}{125}\right)$; find $x$
Answer$ \log _x 25-\log _x 5=2-\log _x\left(\frac{1}{125}\right)$
$ \Rightarrow \log _x 5^2-\log _x 5=2-\log _x\left(\frac{1}{5}\right)^3$
$ \Rightarrow \log _x 5^2-\log _x 5=2-\log _x 5^{-3}$
$ \Rightarrow 2 \log _x 5-\log _x 5=2+3 \log _x 5$
$ \Rightarrow 2 \log _x 5-\log _x 5-3 \log _x 5=2$
$ \Rightarrow-2 \log _x 5=2$
$ \Rightarrow \log _x 5=-1$
$ \Rightarrow x^{-1}=5$
$ \Rightarrow \frac{1}{x}=5$
$ \Rightarrow x=\frac{1}{5}$
View full question & answer→Question 84 Marks
If $\log_{10} 8 = 0.90$; find the value of : $\log_{10}4$
AnswerGiven that $\log _{10} 8=0.90$
$\Rightarrow \log _{10} 2 \times 2 \times 2=0.90$
$\Rightarrow \log _{10} 2^3=0.90$
$\Rightarrow 3 \log _{10} 2=0.90$
$\Rightarrow \log _{10} 2=\frac{0.90}{3}$
$\Rightarrow \log _{10} 2=0.30\ldots(1)$
$\log 4$
$=\log _{10}(2 \times 2)$
$=\log _{10}\left(2^2\right)$
$=2 \log _{10} 2$
$=2(0.30)\ldots[$ from$(1) ]$
$=0.60$
View full question & answer→Question 94 Marks
Solve for $\mathbf{x}: \frac{\log 64}{\log 8}=\log x$
Answer$ \Rightarrow \frac{\log 64}{\log 8}=\log x$
$ \Rightarrow \log x=\frac{\log 64}{\log 8}$
$ \Rightarrow \log x=\frac{\log 2 \times 2 \times 2 \times 2 \times 2 \times 2}{\log 2 \times 2 \times 2}$
$ \Rightarrow \log x=\frac{\log 2^6}{\log 2^3}$
$ \Rightarrow \log x=\frac{6 \log 2}{3 \log 2} \ldots\left[n \log _a m=\log _a m^n\right]$
$ \Rightarrow \log x=\frac{6}{3}$
$ \Rightarrow \log x=2$
$ \Rightarrow \log _{10} x=2$
$ \Rightarrow 10^2=x$
$ \Rightarrow x=10 \times 10$
$ \Rightarrow x=100$
View full question & answer→Question 104 Marks
Evaluate the following without using tables :$\log_{10}8 + \log_{10}25 + 2 \log_{10}3 - \log_{10}18$
AnswerConsider the given expression
$\log _{10} 8+\log _{10} 25+2 \log _{10} 3-\log _{10} 18$
$=\log _{10} 8+\log _{10} 25+\log _{10} 3^2-\log _{10} 18 \ldots .\left[n \log _a m=\log _a m^n\right]$
$=\log _{10} 8+\log _{10} 25+\log _{10} 9-\log _{10} 18$
$=\log _{10} 8 \times 25 \times 9-\log _{10} 18 \ldots .\left[\log _a \mathrm{I}+\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\right.\left.\log _a \operatorname{Imn}\right]$
$=\log _{10} 1800-\log _{10} 18$
$=\log _{10}\left(\frac{1800}{18}\right)$
$=\log _{10} 100 \ldots .\left[\because \log _{10} 100=2\right]$
$=2$
View full question & answer→Question 114 Marks
If $3( \log 5 - \log 3 ) - ( \log 5 - 2 \log 6 ) = 2 - \log x,$ find $x.$
Answer$3(\log 5-\log 3)-(\log 5-2 \log 6)=2-\log x$
$ \Rightarrow 3 \log 5-3 \log 3-\log 5+2 \log (2 \times 3)=2-\log x$
$ \Rightarrow 3 \log 5-3 \log 3-\log 5+2 \log 2+2 \log 3=2-\log x$
$ \Rightarrow 2 \log 5-\log 3+2 \log 2=2-\log x$
$ \Rightarrow 2 \log 5-\log 3+2 \log 2+\log x=2$
$ \Rightarrow \log 5^2-\log 3+\log 2^2+\log x=2$
$ \Rightarrow \log \left(\frac{25 \times 4 \times x}{3}\right)=2$
$ \Rightarrow \log \left(\frac{100 x}{3}\right)=2$
$ \Rightarrow \frac{100 x}{3}=10^2$
$ \Rightarrow \frac{x}{3}=1$
$ \Rightarrow x=3$
View full question & answer→Question 124 Marks
Given $2 \log_{10}x + 1 = \log_{10}250,$ find $:(i) x,(ii) \log_{10}2x$
Answer$(i)$ Consider the given equation:
$2 \log _{10} x+1=\log _{10} 250$
$\Rightarrow \log _{10} x^2+1=\log _{10} 250 \quad\left[\log _a m^n=n \log _a m\right]$
$\Rightarrow \log _{10} x 2+\log _{10} 10=\log _{10} 250\left[\because \log _{10} 10=1\right]$
$\Rightarrow \log _{10}\left(x^2 \times 10\right)=\log _{10} 250 \quad\left[\log _a m+\log _a n=\log _a m n\right]$
$\Rightarrow x^2 \times 10=250$
$\Rightarrow x^2=25$
$\Rightarrow x=\sqrt{25}$
$\Rightarrow x=5$
$(ii) x=5 ($proved above in $(i))$
$\log _{10} 2 x=\log _{10} 2(5)$
$=\log _{10} 10$
$=1\left[\because \log _{10} 10=1\right]$
View full question & answer→Question 134 Marks
If $\log 2 = 0.3010$ and $\log 3 = 0.4771;$ find the value of $: \log 3.6$
AnswerWe know that $\log 2=0.3010$ and $\log 3=0.4771$.
$\log 3.6$
$=\log \frac{36}{10}$
$=\log 36-\log 10 \ldots\left[\log _a\left(\frac{m}{n}\right)=\log _a m-\log _a n\right]$
$=\log 2 \times 2 \times 3 \times 3-1 \ldots[\because \log 10=1]$
$=\log 2 \times 2+\log 3 \times 3-1 \ldots\left[\log _a m n=\log _a m+\log _a n\right]$
$ \left.=\log 2^2+\log 3^2-1 \quad\right]$
$ =2 \log 2+2 \log 3-1 \ldots\left[n \log _a m=\log _a m^n\right]$
$ =2(0.3010)+2(0.4771)-1$
$ =1.5562-1 \ldots[\because \log 2=0.3010$ and $\log 3=0.4771]$
$ =0.5562$
View full question & answer→Question 144 Marks
If $\log 2 = 0.3010$ and $\log 3 = 0.4771 ;$ find the value of $: \log 1.2$
Answer$ \log 2=0.3010$ and $\log 3=0.4771$
$ \log 1.2$
$ =\log \frac{12}{10}$
$ =\log 12-\log 10 \ldots\left[\log a \frac{m}{n}=\log _{\mathrm{a}} \mathrm{m}-\log _{\mathrm{a}} \mathrm{n}\right]$
$ =\log 2 \times 2 \times 3-1 \ldots[\because \log 10=1]$
$ =\log 2 \times 2+\log 3-1 \ldots\left[\log _a m n=\log _a m+\log _a n\right]$
$ =\log 22+\log 3-1$
$ =2 \log 2+\log 3-1 \ldots[ n \log _ a m = \log _ a m n ] $
$ =2(0.3010)+0.4771-1 \ldots[\because \log 2=0.3010$ and $\log 3=0.4771]$
$ =1.0791-1$
$ =0.0791$
$ $
View full question & answer→Question 154 Marks
If $\log_{10}2 = a$ and $\log_{10}3 = b $; express each of the following in terms of $'a'$ and $'b\ '$: $\log 2.25$
Answer$ \log _{10} 2=a$ and $\log _{10} 3=b$
$\log 2.25$
$=\log \frac{225}{100}$
$=\log \frac{25 \times 9}{25 \times 4}$
$=\log \frac{25 \times 9}{25 \times 4}$
$=\log \left(\frac{9}{4}\right)$
$=\log \left(\frac{3}{2}\right)^2$
$=\log \left(\frac{3}{2}\right) \ldots\left[\log _a m=\log _a m^n\right]$
$=2(\log 3-\log 2) \ldots\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$=2(b-a) \ldots\left[\log _{10} 2=a \text { and } \log _{10} 3=b\right]$
$=2 b-2 a $
View full question & answer→Question 164 Marks
Find the logarithm of : $\frac{1}{81}$ to the base $27$
AnswerLet $\log _{27} \frac{1}{81}=x$
$\therefore 27^x=\frac{1}{81}$
$\Rightarrow(3 \times 3 \times 3)^x=\frac{1}{3 \times 3 \times 3 \times 3}$
$\Rightarrow\left(3^3\right)^x=\frac{1}{3^4}$
$\Rightarrow\left(3^3\right)^x=\left(3^4\right)^{-1}$
$\Rightarrow 3^{3 x}=3^{-4}$
$\Rightarrow 3 x=-4 \quad...\dots[$If $a^m=a^n$; then $\left.m=n\right]$
$\Rightarrow x=-\frac{4}{3}$
$\therefore \log _{27} \frac{1}{81}=-\frac{4}{3}$
View full question & answer→Question 174 Marks
Solve for $\mathrm{x}$ and $\mathrm{y}$; if $\mathrm{x}>0$ and $\mathrm{y}>0 ; \log \mathrm{xy}=\log \frac{x}{y}+2 \log 2=2$
Answer$ \log x y=\log \left(\frac{x}{y}\right)+2 \log 2=2$
$ \log x y=2$
$ \Rightarrow \log x y=2 \log 10$
$ \Rightarrow \log x y=\log 10^2$
$ \Rightarrow \log x y=\log 100$
$\therefore x y=100\ldots(1)$
Now consider the equation
$\log \left(\frac{x}{y}\right)+2 \log 2=2$
$ \Rightarrow \log \left(\frac{x}{y}\right)+\log 2^2=2 \log 10$
$ \Rightarrow \log \left(\frac{x}{y}\right)+\log 4=\log 10^2$
$ \Rightarrow \log \left(\frac{x}{y}\right)+\log 4=\log 100$
$ \Rightarrow\left(\frac{x}{y}\right) \times 4=100$
$ \Rightarrow 4 x=100 y$
$ \Rightarrow x=25 y$
$ \Rightarrow x y=25 y \dots y$
$ \Rightarrow x y=25 y^2$
$\Rightarrow 100=25 y^2 \dots...[$ from $(1)]$
$ \Rightarrow \mathrm{y}^2=\frac{100}{25}$
$ \Rightarrow \mathrm{y}^2=4$
$\Rightarrow y=2 \ldots .[\because y>0]$
From $(1),$
$ x y=100$
$ \Rightarrow x \times 2=100$
$ \Rightarrow x=\frac{100}{2}$
$ \Rightarrow x=50 .$
Thus the values of $x$ and $y$ are $x=50$ and $y=2$.
View full question & answer→Question 184 Marks
If $\log \frac{a-b}{2}=\frac{1}{2}(\log a+\log b)$, Show that: $a^2+b^2=6 a b$.
Answer$\log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a+\log b)$
$ \Rightarrow \log \left(\frac{a-b}{2}\right)=\frac{1}{2}(\log a b)$
$ \Rightarrow \log \left(\frac{a-b}{2}\right)=\log (a b)^{\frac{1}{2}}$
$ \Rightarrow\left(\frac{a-b}{2}\right)=(a b)^{\frac{1}{2}}$
Squaring both sides we have,
$\left(\frac{a-b}{2}\right)^2=a b$
$ \Rightarrow \frac{(a-b)^2}{4}=a b$
$ \Rightarrow(a-b)^2=4 a b$
$ \Rightarrow a^2+b^2-2 a b=4 a b$
$ \Rightarrow a^2+b^2=4 a b+2 a b$
$ \Rightarrow a^2+b^2=6 a b .$
View full question & answer→Question 194 Marks
If $a^2= \log x, b^3= \log y$ and $3a^2- 2b^3 = 6 \log z,$ express $y$ in terms of $x$ and $z$ .
AnswerGiven that
$a^2=\log x_t b^3=\log y$ and $3 a^2-2 b^3=6 \log z$
Consider the equation,
$3 a^2-2 b^3=6 \log z$
$ \Rightarrow 3 \log x-2 \log y=6 \log z$
$ \Rightarrow \log x^3-\log y^2=\log z^6$
$ \Rightarrow \log \left(\frac{x^3}{y^2}\right)=\log z^6$
$ \Rightarrow \frac{x^3}{y^2}=z^6$
$ \Rightarrow \frac{x^3}{z^6}=y^2$
$ \Rightarrow y^2=\frac{x^3}{z^6}$
$ \Rightarrow y=\left(\frac{x^3}{z^6}\right)^{\frac{1}{2}}$
$ \Rightarrow y=\left(\frac{x^{\frac{3}{2}}}{z^{\frac{6}{2}}}\right)$
$ \Rightarrow y=\frac{x^{\frac{3}{2}}}{z^3}$
View full question & answer→Question 204 Marks
If $x = 1 + \log 2 - \log 5, y = 2 \log3$ and $z = \log a - \log 5;$ find the value ofaif $x + y = 2z.$
AnswerGiven that
$x=1+\log 2-\log 5,$
$ y=2 \log 3$ and
$ z=\log a-\log 5$
Consider
$x=1+\log 2-\log 5$
$ =\log 10+\log 2-\log 5$
$ =\log (10 \times 2)-\log 5$
$ =\log 20-\log 5$
$ =\log \frac{20}{5}$
$=\log 4\ldots .(1)$
We have
$y=2 \log 3$
$ =\log 3^2$
$=\log 9\ldots .(2)$
Also we have
$z=\log a-\log 5$
$=\log \frac{a}{5}\ldots (3)$
Given that $x+y=2 z$
$\therefore$ Subsitute the values of $x_1 y_1$ and $z$. from $(1), (2)$ and $(3),$ We have
$\Rightarrow \log 4+\log 9=2 \log \frac{a}{5}$
$ \Rightarrow \log 4+\log 9=\log \left(\frac{a}{5}\right)^2$
$ \Rightarrow \log 4+\log 9=\log \left(\frac{a^2}{25}\right)$
$ \Rightarrow \log (4 \times \log 9)=\log \left(\frac{a^2}{25}\right)$
$ \Rightarrow \log 36=\log \left(\frac{a^2}{25}\right)$
$\Rightarrow \frac{a^2}{25}=36$
$ \Rightarrow a^2=36 \times 25$
$ \Rightarrow a^2=900$
$ \Rightarrow a=30 .$
View full question & answer→Question 214 Marks
If $\log _2(x+y)=\log _3(x-y)=\frac{\log 25}{\log 0.2}$, find the values of $x$ and $y$.
Answer$ \log _2(x+y)=\frac{\log 25}{\log 0.2}$
$ \Rightarrow \log _2(x+y)=\log _{0.2} 25$
$ \Rightarrow \log _2(x+y)=\log _{\frac{2}{10}} 25$
$ \Rightarrow \log _2(x+y)=\log _5^{-1} 5^2$
$ \Rightarrow \log _2(x+y)=-2 \log _5 5$
$ \Rightarrow \log _9(x+y)=-2$
$\Rightarrow x+y=2^{-2} \dots...[$ Removing logarithm$ ]$
$\Rightarrow x+y=\frac{1}{4}\ldots .(1)$
$ \log _3(x-y)=\frac{\log 25}{\log 0.2}$
$ \Rightarrow \log _3(x-y)=\log _{0.2} 25$
$ \Rightarrow \log _3(x-y)=\log _{\frac{2}{10}} 25$
$ \Rightarrow \log _3(x-y)=\log _5^{-1} 5^2$
$ \Rightarrow \log _3(x-y)=-2 \log _5 5$
$ \Rightarrow \log _3(x-y)=-2$
$\Rightarrow x-y=3^{-2}\dots ...[$ Removing logarithm $]$
$\Rightarrow x-y=\frac{1}{9}\ldots(2)$
Solving $(1)$ and $(2),$ We get
$x=\frac{13}{72}, y=\frac{5}{72}$
View full question & answer→Question 224 Marks
If $\log_{10} 8 = 0.90;$ find the value of : $\log 0.125$
AnswerGiven that $\log _{10} 8=0.90$
$\Rightarrow \log _{10} 2 \times 2 \times 2=0.90$
$ \Rightarrow \log _{10} 2^3=0.90$
$ \Rightarrow 3 \log _{10} 2=0.90$
$ \Rightarrow \log _{10} 2=\frac{0.90}{3}$
$\Rightarrow \log _{10} 2=0.30\ldots(1)$
$ \log 0.125$
$ =\log _{10} \frac{125}{1000}$
$ =\log _{10} \frac{1}{8}$
$ =\log _{10}\left(\frac{1}{2 \times 2 \times 2}\right)$
$ =\log _{10}\left(\frac{1}{2^3}\right)$
$ =\log _{10} 2^{-3}$
$ =-3 \times(0.30) \quad[$from $(1)]$
$ =-0.9$
View full question & answer→Question 234 Marks
If $ \log_{10} 8 = 0.90;$ find the value of : $\log\sqrt{32}$
AnswerGiven that $\log _{10} 8=0.90$
$ \Rightarrow \log _{10} 2 \times 2 \times 2=0.90$
$ \Rightarrow \log _{10} 2^3=0.90$
$ \Rightarrow 3 \log _{10} 2=0.90$
$ \Rightarrow \log _{10} 2=\frac{0.90}{3}$
$\Rightarrow \log _{10} 2=0.30\ldots(1)$
$ \log \sqrt{32}$
$ =\log _{10}(32)^{\frac{1}{2}}$
$ =\frac{1}{2} \log _{10}(32)$
$ =\frac{1}{2} \log _{10}(2 \times 2 \times 2 \times 2 \times 2)$
$ =\frac{1}{2} \log _{10}\left(2^5\right)$
$ =\frac{1}{2} \times 5 \log _{10} 2$
$=\frac{1}{2} \times 5(0.30)$
$[$from$(1)]$
$ =5 \times 0.15$
$ =0.75$
View full question & answer→Question 244 Marks
Solve for $x : \log (x + 5) + \log (x - 5) = 4 \log 2 + 2 \log 3$
Answer$\log (x+5)+\log (x-5)=4 \log 2+2 \log 3$
$\Rightarrow \log (x+5)(x-5)=4 \log 2+2 \log 3 \ldots\left[\log _a m+\log _a n+\log _a m n\right]$
$\Rightarrow \log \left(x^2-25\right)=\log 2^4+\log 3^2 \ldots\left[n \log _a m=\log _a m^n\right]$
$ \Rightarrow \log \left(x^2-25\right)=\log 16+\log 9$
$ \Rightarrow \log \left(x^2-25\right)=\log 16 \times 9 \ldots\left[\log _a m+\log _a n+\log _a m n\right]$
$ \Rightarrow \log \left(x^2-25\right)=\log 144$
$ \Rightarrow x^2-25=144$
$ \Rightarrow x^2=144+25$
$ \Rightarrow x^2=169$
$ \Rightarrow x= \pm \sqrt{169}$
$ \Rightarrow x= \pm \sqrt{13^2}$
$ \Rightarrow x= \pm 13$
View full question & answer→Question 254 Marks
Find $x$, if $: x-\log 48+3 \log 2=\frac{1}{3} \log 125-\log 3$.
AnswerConsider the given equation
$x-\log 48+3 \log 2=\frac{1}{3} \log 125-\log 3$
$ \Rightarrow x=\frac{1}{3} \log 125-\log 3+\log 48-3 \log 2$
$ \Rightarrow x=$
$ \log (125)^{\frac{1}{3}}-\log 3+\log 48-\log 2^3 \ldots .\left[n \log _a m=\log _a m^n\right]$
$ \Rightarrow x=\log (5 \times 5 \times 5)^{\frac{1}{3}}-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log \left(5^3\right)^{\frac{1}{3}}-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log 5-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log 5+\log 48-\log 3-\log 8$
$ \Rightarrow x=(\log 5+\log 48)-(\log 3+\log 8)$
$ \Rightarrow \mathrm{x}=(\log 5 \times 48)-(\log 3 \times 8) \ldots .\left[\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \mathrm{mn}\right]$
$ \Rightarrow \mathrm{x}=\log \frac{5 \times 48}{3 \times 8} \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ \Rightarrow \mathrm{x}=\log \frac{5 \times 6 \times 8}{3 \times 8}$
$ \Rightarrow \mathrm{x}=\log 10$
$ \Rightarrow \mathrm{x}=1$
View full question & answer→Question 264 Marks
Prove that : $2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log 2$
AnswerWe need to prove that
$2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log 2$
$ \text { LHS }=2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}$
$ =\log \left(\frac{15}{18}\right)^2-\log \left(\frac{25}{162}\right)+\log \left(\frac{4}{9}\right) \ldots .\left[\mathrm{n} \log _{\mathrm{a}} \mathrm{m}=\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}\right.$
$ =\log \left[\left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right)\right]-\log \frac{25}{162}+\log \frac{4}{9}$
$ =\log \left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right) \times\left(\frac{4}{9}\right)-\log \left(\frac{25}{162}\right) \ldots . .\left[\log _a m+\log _a n=\log _a(m n)\right]$
$ =\log \frac{\left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right) \times \frac{4}{9}}{\frac{25}{162}} \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log \left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right) \times \frac{4}{9} \times \frac{162}{25}$
$ =\log \frac{72}{36}$
$ =\log 2$
$ =\text { R.H.S. }$
View full question & answer→Question 274 Marks
If $x=(100)^a, y=(10000)^b$ and $z=(10)^c$, find $\log \frac{10 \sqrt{y}}{x^2 z^3}$ in terms of $a, \mathrm{b}$ and $\mathrm{c}$.
Answer$ x=(100)^a, y=(10000)^b$ and $z=(10)^c$
$ \Rightarrow \log x=a \log 100, \log y=b \log 10000$ and $\log z=c \log 10$
$ \log \frac{10 \sqrt{y}}{x^2 z^3}=\log 10 \sqrt{y}-\log \left(x^2 z^3\right)$
$ =\log \left(10 y^{1 / 2}\right)-\log x^2-\log z^3$
$ =\log 10+\log y^{1 / 2}-\log x^2-\log z^3$
$ =\log 10+\frac{1}{2} \log y-2 \log x-3 \log z$
$=1+\frac{1}{2} \log (10000)^b-2 \log (100)^a-3 \log (10)^c \ldots . .($ Since $\log 10= 1)$
$ =1+\frac{b}{2} \log (10)^4-a \log (10)^2-3 c \log 10$
$ =1+\frac{b}{2} \times 4 \log 10-2 \times 2 a \log 10-3 c \log 10$
$ =1+2 b-4 a-3 c$
View full question & answer→Question 284 Marks
Express in terms of $\log 2$ and $\log 3 :\log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}$
Answer$ \log \frac{75}{16}-2 \log \frac{5}{9}+\log \frac{32}{243}$
$ =\log \frac{75}{16}-\log \left(\frac{5}{9}\right)^2+\log \frac{32}{243}$
$ =\log \frac{75}{16}-\log \left(\frac{5}{9} \times \frac{5}{9}\right)+\log \frac{32}{243}$
$ =\log \frac{75}{16}-\log \frac{25}{81}+\log \frac{32}{243}$
$ =\log \left(\frac{\frac{75}{16}}{\frac{25}{81}}\right) \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log \left(\frac{75}{16}\right) \times\left(\frac{81}{25}\right)+\log \left(\frac{32}{243}\right)$
$ =\log \frac{3 \times 25}{16} \times \frac{81}{25}+\log \frac{32}{243}$
$ =\log \frac{243}{16}+\log \frac{32}{243}$
$ =\log \left(\frac{243}{16} \times \frac{32}{243}\right) \ldots \ldots\left[\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \mathrm{mn}\right]$
$ =\log \frac{32}{16}$
$ =\log 2$
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