First of all, rewrite
[tex]4^{x+2}=4^x\cdot 4^2=16\cdot 4^x[/tex]
And divide both sides by 16 to get
[tex]4^x=\dfrac{12}{16}=\dfrac{3}{4}[/tex]
The solution is
[tex]x=\log_4\left(\dfrac{3}{4}\right)[/tex]
Which, using the suggested formula, we can compute as
[tex]\dfrac{\log\left(\frac{3}{4}\right)}{\log(4)}[/tex]
Which evaluates to about - 0.20752..
