Question

Solve ${\log }_2({25}^{x+3}-1)=2+{\log }_2(5^{x+3}+1)$

Answer 1

${\log }_2\left({25}^{x+3}-1\right)=2+{\log }_2\left(5^{x+1}+1\right)$

$={\log }_2\left({25}^{x+3}-1\right)=2{\log }_2^2+{\log }_2\left(5^{x+3}+1\right)$

$={\log }_2\left({25}^{x+3}-1\right)={\log }_{2}4\left(5^{x+3}+1\right)$

$={25}^{x+3}-1=4\left(5^{x+3}+1\right)$

$=5^{2\left(x+3\right)}-1=4\times 5^{x+3}+4$

$Let{5}^{x+3}=y$

$y^2-1=4y+4$

$y^2-4y-5=0$

$y^2+y-5y-5=0$

$y\left(y+1\right)-5\left(y+1\right)=0$

$\left(y+1\right)\left(y-5\right)=0$

$y=1,5$

$5^{x+3}\ne 1$

$5^{x+3}=5$