Question

Find the value of $x$, if ${\log }_2({25}^{x+3}-1)=2+{\log }_2(5^{x+3}+1)$

Answer 1

${\log }_2({25}^{x+3}-1)=2+{\log }_2(5^{x+3}+1)$
${\log }_2\left(\frac{{25}^{x+3}-1}{5^{x+3}+1}\right)=2[\because \log a-\log b=\log \frac{a}{b}]$
$\Rightarrow \frac{5^{2(x+3)}-1}{5^{x+3}+1}=4$

$\Rightarrow 4(5^{x+3}+1)=5^{2(x+3)}-1$
$\Rightarrow 5^{2(x+3)}-{4.5}^{x+3}-5=0$
Substitute $5^{x+3}=t$
$\Rightarrow t^2-4t-5=0$
$\Rightarrow (t-5)(t+1)=0$
$\Rightarrow t=5,t=-1$
$\Rightarrow 5^{x+3}=5,5^{x+3}=-1$
$\Rightarrow x+3=1$
$\Rightarrow x=-2$
Now, from given equation, it follows that for logarithm to be defined
${25}^{x+3}>1,5^{x+3}>-1$
$x=-2$ satisfies above inequalities
Hence, $x=-2$ is a solution of the given equation