[5 marks sum]

Question 15 Marks

Simplify the following:$3 \log \frac{32}{27}+5 \log \frac{125}{24}-3 \log \frac{625}{243}+\log \frac{2}{75}$

Answer

$3 \log \frac{32}{27}+5 \log \frac{125}{24}-3 \log \frac{625}{243}+\log \frac{2}{75} $
$ =3 \log \frac{2^5}{3^3}+5 \log \frac{5^3}{2^3 \times 3}-3 \log \frac{5^4}{2 \times 3^4}+\log \frac{2}{3 \times 5^2}$
$ =3 \log 2^5-3 \log 3^3+5 \log 5^3-5 \log 2^3-5 \log 3-3 \log 5^4+3 \log 2+3 \log 3^4+\operatorname{loh} 2-\log 3- $
$\log 5^2$
$ =3 \times 5 \log 2-3 \times 3 \log 3+5 \times 3 \log 5-5 \times 3 \log 2-5 \log 3-3 \times 4 \log 5+3 \log 2+3 \times 4 \log 3+ $
$ \log 2-\log 3-2 \log 5 $
$=15 \log 2-9 \log 3+15 \log 5-15 \log 2-5 \log 3-12 \log 5+3 \log 2+12 \log 3+\log 2-\log 3- $
$ 2 \log 5$
$=\log 5+\log 2 $
$ =\log (5 \times 2) $
$ =\log 10 $
$ =1$

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Question 25 Marks

Prove that: $\frac{1}{\log _8 36}+\frac{1}{\log _9 36}+\frac{1}{\log _{18} 36}=2$

Answer

$=\frac{1}{\log _8 36}+\frac{1}{\log _9 36}+\frac{1}{\log _{18} 36} $
$=\log _{36} 8+\log _{36} 9+\log _{36} 18 $
$=\frac{\log 8}{\log 36}+\frac{\log 9}{\log 36}+\frac{\log 18}{\log 36} $
$=\frac{1}{\log 36}(\log 8+\log 9+\log 18) $
$=\frac{1}{\log 36}\left(\log 2^3+\log 3^2+\log \left(2 \times 3^2\right)\right) $
$=\frac{1}{\log \left(2^2 \times 3^2\right)}\left(\log 2^3+\log 3^2+\log 2+\log 3^2\right) $
$=\frac{1}{\log \left(2^2 \times 3^2\right)}(3 \log 2+2 \log 3+\log 2+\log 3) $
$=\frac{1}{2 \log 2+2 \log 3}(4 \log 2+4 \log 3) $
$=\frac{4}{2(\log 2+\log 3)}(\log 2+\log 3) $
$=2 $
$=\text { R.H.S. }$
Hence proved.

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Question 35 Marks

Prove that: $\frac{1}{\log _2 30}+\frac{1}{\log _3 30}+\frac{1}{\log _5 30}=1$

Answer

$=\frac{1}{\log _2 30}+\frac{1}{\log _3 30}+\frac{1}{\log _5 30} $
$=\frac{1}{\frac{\log 30}{\log 2}}+\frac{1}{\frac{\log 30}{\log 3}}+\frac{1}{\frac{\log 30}{\log 5}} $
$=\frac{\log 2}{\log 30}+\frac{\log 3}{\log 30}+\frac{\log 5}{\log 30} $
$=\frac{1}{\log 30}(\log 2+\log 3+\log 5) $
$=\frac{1}{\log (2 \times 3 \times 5)}(\log 2+\log 3+\log 5) $
$=\frac{(\log 2+\log 3+\log 5)}{(\log 2+\log 3+\log 5)} $
$=1 $
$=\text { L.H.S. }$
Hence proved.

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Question 45 Marks

Find the value of:$\frac{\log \sqrt{27}+\log 8+\log \sqrt{1000}}{\log 120}$

Answer

$\frac{\log \sqrt{27}+\log 8+\log \sqrt{1000}}{\log 120} $
$ =\frac{\log (27)^{\frac{1}{2}}+\log 2^3+\log 1000^{\frac{1}{2}}}{\log \left(3 \times 2^2 \times 10\right)} $
$ =\frac{\log (3)^{3 \times \frac{1}{2}}+\log 2^3+\log (10)^{3 \times \frac{1}{2}}}{\log 3+\log 2^{\wedge} 2+\log 10} $
$ =\frac{\frac{3}{2} \log 3+3 \log 2+\frac{3}{2} \log (10)}{\log 3+2 \log 2+\log 10} $
$ =\frac{\frac{3}{2} \log 3+\frac{3}{2}(2 \log 2)+\frac{3}{2}(1)}{\log 3+2 \log 2+1} $
$ =\frac{\frac{3}{2}[\log 3+2 \log 2+1]}{\log 3+2 \log 2+1}$
$ =\frac{3}{2} .$

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Question 55 Marks

If $2 \log x + 1 = \log$ $360$, find: $\log (3 x^2 - 8)$

Answer

$\log \left(3 x^2-8\right) $
$2 \log x+1=\log 360 $
$\Rightarrow \log x^2+\log 10=\log 360 $
$\Rightarrow \log \left(10 x^2\right)=\log 360 $
$\Rightarrow 10 x^2=360 $
$\Rightarrow x^2=\frac{360}{10}=36 $
$\Rightarrow x=\sqrt{36}= \pm 6$
As negative value is rejected,
$\therefore x =6 $
$\therefore \log \left(3 x ^2-8\right) $
$=\log \left\{3(6)^2-8\right\} $
$=\log (108-8) $
$=\log 100 $
$=\log 10^2 $
$=2 \log 10 $
$=2 \times 1$
$= 2.$

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Question 65 Marks

Find $x$ and $y$, if $\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}$

Answer

$\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}=\frac{\log 64}{\log y}$
Considering the first equality
$\frac{\log x}{\log 5}=\frac{\log 36}{\log 6}$
$\Rightarrow \frac{\log x}{\log 5}=\frac{\log 6^2}{\log 6}=\frac{2 \log 6}{\log 6}=2 $
$\Rightarrow \log x=2 \log 5=\log 5^2=\log 25$
$\therefore x =25$
Considering the second equality
$\frac{\log 36}{\log 6}=\frac{\log 64}{\log y} $
$\Rightarrow \frac{\log 6^2}{\log 6}=\frac{2 \log 6}{\log 6}=2=\frac{\log 8}{\log y} $
$\Rightarrow \log y=\frac{\log 64}{2}=\frac{\log 8^2}{2}=\frac{2 \log 8}{2}=\log 8 $
$\therefore y=8$

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Question 75 Marks

If $\log 16 = a, \log 9 = b$ and $\log 5 = c,$ evaluate the following in terms of $a, b, c: \log 2.25$

Answer

$\log 16=a, \log 9=b$ and $\log 5=c$
$\log 4^2=a, \log 3^2=b$ and $\log 5=c$
$2 \log 4=a, 2 \log 3=b$ and $\log 5=c$
$\log 4=\frac{a}{2}, \log 3=\frac{b}{2}$ and $\log 5=c$
Consider,$ \log 2.25=\log \left(\frac{225}{100}\right)$
$=\log 225-\log 100$
$=\log \left(3^2 \times 5^2\right)-\log \left(4 \times 5^2\right)$
$=\log 3^2+\log 5^2-\left(\log 4+\log 5^2\right)$
$=\log 3^2+\log 5^2-\log 4-\log 5^2$
$=2 \log 3-\log 4$
$=2\left(\frac{b}{2}\right)-\frac{a}{2}$
$=\frac{2 b-a}{2} .$

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MATHEMATICS [5 marks sum]

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