Question

Solve the following equations
$\log (2^x+x-41)$ = $x(1-\log 5)$.

Answer 1

$\log (2^x+x-41)=x(1-\log 5)$

Use : $\log 10=1$

$\Rightarrow \log (2^x+x-41)=x(\log 10-\log 5)$

use $\log \left(\frac{a}{b}\right)=\log a-\log b$

$\Rightarrow \log (2^x+x-41)=x\left(\log \frac{10}{5}\right)$

$\Rightarrow \log (2^x+x-41)=x\log 2$

Use : $a\log b=\log \left(b^a\right)$

$\Rightarrow \log (2^x+x-41)=\log \left(2^x\right)$

Remove $\log$ from both sides

$\Rightarrow 2^x+x-41=2^x$

$\Rightarrow x-41=0$

$\Rightarrow x=41$.