Answer
$ x=\dfrac{1}{2}$ or, $x=-3$
Work Step by Step
Recall the rule that when $y=\log_a b$, then we have: $a^y=b$
Applying this rule, we obtain:
$2^{3}=(2^2)^{x^2} 2^{5x} \\ 2^3 =2^{2x^2} (2^{5x}) \\ 2^3=2^{2x^2+5x}$
The base is the same, so both sides can be equal when the exponents are equal.
$2x^2+5x=3 \\ 2x^2+5x-3=0$
This yields a quadratic equation whose factors are: $(2x-1)(x+3)=0$
By the zero property, we have:
$2x-1=0 \implies x=\dfrac{1}{2}$
and
$x+3 =0 \implies x=-3$
Therefore, $ x=\dfrac{1}{2}$ or, $x=-3$
