[5 marks sum]

Question 15 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 x 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 $y=2$.

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Question 25 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 .$

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Question 35 Marks

If $a^2=\log x, b^3=\log y$ and $3 a^2-2 b^3=6 \log z$, express $y$ in terms of $x$ and $z$.

Answer

Given that
$a^2=\log x_t b^3=\log y \text { and } 3 a^2-2 b^3=6 \log z$
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}$

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Question 45 Marks

If $x = 1 + \log 2 - \log 5, y = 2 \log3$ and $z = \log a - \log 5;$ find the value ofaif $x + y = 2z.$

Answer

Given 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 .$

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Question 55 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
$x=\frac{13}{72}, y=\frac{5}{72}$

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Question 65 Marks

If $\log_{10}8 = 0.90$; find the value of : $\log 0.125$

Answer

Given 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$

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Question 75 Marks

If $\log _{10} 8=0 . 9 0$; find the value of : $\log \sqrt{32 } $

Answer

Given 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$

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Question 85 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$

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Question 95 Marks

Find $x$, if $: x-\log 48+3 \log 2=\frac{1}{3} \log 125-\log 3$.

Answer

Consider 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$

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Question 105 Marks

Prove that : $2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log 2$

Answer

We 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. }$

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Question 115 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$

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Question 125 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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MATHEMATICS [5 marks sum]

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