Question

Evaluate : $4\cdot 2^{2x}-6^x={18.3}^{2x}$

Answer 1

⇒${4.2}^{2x}-6^x={18.3}^{2x}$

⇒$4(\frac{2}{3}{)}^{2x}-(\frac{2}{3}{)}^x=18$

Let $(\frac{2}{3}{)}^x=t$

⇒ $4t^2-t-18=0$

On solving we get $t=\frac{9}{4}$ ( $t=-2$ is rejected since $(\frac{2}{3}{)}^x$ is positive )

⇒So $x=-2$