Question

Solve
${\log }_{10}\left(\frac{1}{2^x+1-1}\right)=x[{\log }_{10}5-1]$

Answer 1

${\log }_{10}\left(\frac{1}{2^n+1-1}\right)=x[{\log }_{10}5-1]$

${\log }_{10}\frac{1}{2^x}=x{\log }_{10}5-x$

$\because \log \frac{x}{y}=\log x-\log y$

$\Rightarrow {\log }_{10}1-{\log }_{10}{2}^x={\log }_{10}{5}^x-x$

$\because {\log }_{10}1=0$

$\Rightarrow -{\log }_{10}{2}^x={\log }_{10}{5}^x-x$

$\Rightarrow x={\log }_{10}{5}^x+{\log }_{10}{2}^x$ $\because \log \left(ab\right)=\log a+\log b$

$\Rightarrow x={\log }_{10}(5^x\times 2^x)$

$\Rightarrow x=\frac{{\log }_e(5^x\times 2^x)}{{\log }_{e}10}$ $\because {\log }_{10}x=\frac{\log x}{\log 10}$

$\Rightarrow x={\log }_e(5^x\times 2^x)$ $\because {\log }_{10}=1$

$\Rightarrow e^x=5^x\times 2^x$

Hence, this is the answer.