Question

Simplify : $f\left(x\right)=\frac{log100-log(0.01-x)}{3x}$

Answer 1

We have,

$f\left(x\right)=\frac{\log 100-\log (0.01-x)}{3x}$

$=\frac{\log \frac{100}{(0.01-x)}}{3x}\therefore \log a-\log b=\log \frac{a}{b}$

$=\frac{1}{3x}\log \frac{100}{(0.01-x)}$

$=\frac{1}{3x}\log \frac{100}{(\frac{1}{100}-x)}$

$=\frac{1}{3x}\log \frac{100}{\frac{1-100x}{100}}$

$=\frac{1}{3x}\log \frac{10000}{1-100x}$

Hence, this is the answer.