Logarithm inequality $\log (25 + 5^{2x}) > x(1-\log 2) + \log 2 + \log13$

Question

known $\log (25 + 5^{2x}) > x(1-\log 2) + \log 2 + \log13$

Find range of $x$ Options : $x<0 \cup x>2$

$0<x<2$

$x>2$

$x\leq 0 \cup x>2$ By substituting i find $x=0,x=1$ not sufficient. So the answer is$x<0 \cup x>2$ But how to solve it manually since i tried but not find it

Some hints anyone?

Answer 1

Assuming the base $=10$

and using $\log a+\log b=\log(ab)$ and $\log(a/b)=\log a-\log b$

$$(25+25^x)/26>5^x$$

Set $5^x=y$ to find $$0<y^2-26y+25=(y-1)(y-25)$$

So either $y<1$ or $y >25$

Answer 2

Hints

1. Differentiate the function $f(x)=\log (25 + 5^{2x}) - x(1-\log 2) - \log 2 - \log13$. $$f'(x)\begin{cases}\leq0 &\text{ for } x\leq1\\ >0&\text{ for } x>1.\end{cases}$$
2. $f(0)=f(2)=0$