Question

Solve
$2^{2x+2}-6^x-{2.3}^{2x+2}=0$

Answer 1

We have,

$2^{2x+2}-6^x-{2.3}^{2x+2}=0$

now,

$\Rightarrow 2^{2(x+1)}-6^x-{2.3}^{2(x+1)}=0$

$\Rightarrow 4^{(x+1)}-6^x-{2.9}^{(x+1)}=0$

$\Rightarrow 4^x.4-2^x{.3}^x-2.\left(9^x.9\right)=0$

$\Rightarrow 4.(2^x{)}^2-2^x{.3}^x-18(3^{x^2})=0$

on divide $\left(2^{x^2}\right)$ and we get,

$\Rightarrow \frac{4.(2^x{)}^2}{(2^x{)}^2}-\frac{2^x{.3}^x}{(2^x{)}^2}-18{\left[{\left(\frac{3}{2}\right)}^x\right]}^2=0$

$\Rightarrow 4-{\left(\frac{3}{2}\right)}^x-18{\left[{\left(\frac{3}{2}\right)}^x\right]}^2=0$

let, ${\left(\frac{3}{2}\right)}^x=y----\left(1\right)$

now,

$\Rightarrow 4-y-18y^2=0$

on factorize

$\Rightarrow 4-(9-8)y-18y^2=0$

$\Rightarrow 4-9y+8y-18y^2=0$

$\Rightarrow 1(4-9y)+2y(4-9y)=0$

$\Rightarrow (4-9y)(1+2y)=0$

$\Rightarrow 4-9y=0,1+2y=0$

$\Rightarrow y=\frac{4}{9}\to$ (choose), $y=\frac{-1}{2}$ (not choose)

using equation $\left(1\right)$

${\left(\frac{3}{2}\right)}^x=\frac{4}{9}$

$\Rightarrow {\left(\frac{3}{2}\right)}^x={\left(\frac{2}{3}\right)}^2$

on reciprocal

$\Rightarrow {\left(\frac{3}{2}\right)}^x={\left(\frac{3}{2}\right)}^{-2}$

then, comparing,

$x=-2$

Hence, this is the answer.