Question

$f\left(x\right)=\frac{log\frac{1}{2}+log(2+4x)}{x}$ for $x\ne 0$ at x =2

Answer 1

$f\left(x\right)=\frac{\log \frac{1}{2}+\log (2+4x)}{x}$ for $x\pm 0$ at $x=8$

$\Rightarrow f\left(x\right)=\frac{\log \frac{1}{2}+\log (2+4x)}{x}$

$=\frac{\log \frac{1}{2}+\log (2+8)}{2}$

$=\frac{\log \frac{1}{2}+\log \left(10\right)}{2}$

$=\frac{\log 1+\log 2+1}{2}$

$=\frac{-\log 2+1}{2}$