[3 marks sum]

Question 513 Marks

Find the logarithm of :$ 0.125$ to the base $2$

Answer

$ \text { Let } \log _2 0.125=\mathrm{x}$
$ \therefore 2^x=0.125$
$ \Rightarrow 2^x=\frac{125}{1000}$
$ \Rightarrow 2^x=\frac{1}{8}$
$ \Rightarrow 2^x=8^{-1}$
$ \Rightarrow 2^x=(2 \times 2 \times 2)^{-1}$
$ \Rightarrow 2^x=\left(2^3\right)^{-1}$
$ \Rightarrow 2^x=2^{-3}$
$ \Rightarrow x=-3 \ldots . .\left[\right.$ If $a^m=a^n$; then $\left.m=n\right]$
$ \therefore \log _2 0.125=-3$

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Question 523 Marks

Find the logarithm of : $32$ to the base $4$

Answer

Let $\log _4 32=\mathrm{x}$
$ \therefore 4^{x}=32$
$ \Rightarrow\left(2^2\right)^{x}=2 \times 2 \times 2 \times 2 \times 2$
$ \Rightarrow 2^2 {x}=2^5$
$\Rightarrow 2 x=5 ..... \left[\right.$ If $a^m=a^n$; then $\left.m=n\right]$
$ \Rightarrow x=\frac{5}{2}$
$ \therefore \log _4 32=\frac{5}{2}$

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Question 533 Marks

Find the logarithm of : $0.001$ to the base $10$

Answer

$ \text { Let } \log _{10} 0.001=\mathrm{x}$
$ \therefore 10^{\mathrm{x}}=0.001$
$ \Rightarrow 10^{\mathrm{x}}=\frac{1}{1000}$
$ \Rightarrow 10^{\mathrm{x}}=\frac{1}{10^3}$
$ \Rightarrow 10^{\mathrm{x}}=10^{-3}$
$ \Rightarrow x=-3 \ldots . .\left[\right.$ If $a^m=a^n$; then $\left.m=n\right]$
$\therefore \log _{10} 0.001=-3$

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Question 543 Marks

Solve for $x :\log_{10}x = -2.$

Answer

$ \log _{10} x=-2$
$ \Rightarrow 10^{-2}=x $
$ \Rightarrow x=10^{-2}$
$\Rightarrow x=\frac{1}{10^2} $
$ \Rightarrow x=\frac{1}{100} $
$\Rightarrow x=0.01$

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Question 553 Marks

Solve for $x : \log( x - 1 ) + \log (x + 1 ) = \log_21.$

Answer

$ \log (x-1)+\log (x+1)=\log _2 1$
$\Rightarrow \log (x-1)+\log (x+1)=0$
$\Rightarrow \log [(x-1)(x+1)]=0$
$\Rightarrow(x-1)(x+1)=1 \ldots .($ Since $\log 1=0)$
$\Rightarrow x^2-1=1$
$\Rightarrow x^2=2$
$\Rightarrow x= \pm \sqrt{2}$
$-\sqrt{2}$ can not be possible, since $\log$ of a negative number is not defined.
So,$x=\sqrt{2} \text {. }$

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Question 563 Marks

If $\log_2x = a$ and $\log_3 y = a$, write $72^a$ in terms of $x$ and $y.$

Answer

Given that :
$\log_2x = a$ and $\log_3y = a$
$\Rightarrow 2^a= x$ and $3^a = y ....[ Q \log_am = n \Rightarrow a^n = m ]$
Now prime factorization of $72$ is
$72 = 2 \times 2 \times 2 \times 3 \times 3$
Hence,
$(72)^a = (2 \times 2 \times 2 \times 3 \times 3)^a$
$= (2^3\times 3^2)^a$
$= 2^{3a}\times 3^{2a}$
$= (2^a)^3 \times (3^a)^2 ....[$ as $2^a = x , 3^a = y ]$
$= x^3y^2$

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MATHEMATICS [3 marks sum]