Question

Solve the following equation:
$\log (5^{x-2}+1)=x+\log 13-2\log 5+(1-x)\log 2$

Answer 1

$\log \left(\frac{5^x+25}{5^2}\right)=x+\log 13-2\log 5+(1-x)\log 2\log (5^x+25)-2\log 5=x+\log 13-2\log 5+(1-x)\log 2\log (5^x+25)=x+\log 13+(1-x)\log 2\log (5^x+25)-\log 13=(1-\log 2)x+\log 2\log (5^x+25)-\log 13-\log 2=(1-\log 2)x\log \left(\frac{5^x+25}{26}\right)=\log \left(5\right)x{5}^x+25=\left(5^x\right)2625=25\left(5^x\right)5^x=1x=0$