Question

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

Answer 1

Given that ${\log }_2({25}^{x+3}-1)=2+lo{g}_2(5^{x+3}+1)$

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

$⟹{\log }_2\left(\frac{5^{2(x+3)}-1}{5^{(x+3)}+1}\right)=2$ (since, $\log (a-b)=\log \left(\frac{a}{b}\right))$

$⟹\frac{5^{2(x+3)}-1}{5^{(x+3)}+1}=2^2$ (since, ${\log }_{a}x=b⟹a^b=x$)

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

$⟹(5^{(x+3)}{)}^2-1-4(5^{(x+3)})-4=0$

Let $t=5^{(x+3)}$

$⟹t^2-4t-5=0$

$⟹t^2-5t+t-5=0$

$⟹t(t-5)+1(t-5)=0$

$⟹(t-5)(t+1)=0$

$⟹t=5$ or $t=-1$

But, $5^{(x+3)}=-1$ is not possible.

Therefore, $5^{(x+3)}=5$

Bases are equal. Therefore, powers are equated.

$⟹x+3=1⟹x=-2$