Question

Solve the following equation for $x$ :

ii) $3^{2x+4}+1=2\times 3^{x+2}$

Answer 1

$3^{2x+4}+1=2\times 3^{x+2}$

$\Rightarrow 3^{2x}\times 3^4+1=2\times 3^x\times 3^2$

$[\because a^{m+n}=a^m\times a^n]$

$\Rightarrow 3^{2x}\times 81+1=18\times 3^x$

$\Rightarrow (3^x{)}^2\times 81-18\times 3^x+1=0$

$[\because (a^m{)}^n=a^{mn}]$

let $3^x=t\cdots \left(i\right)$

$\Rightarrow 81(t{)}^2-18t+1=0$

$\Rightarrow 81(t{)}^2-9t-9t+1=0$

$\Rightarrow 9t(9t-1)-1(9t-1)=0$

$\Rightarrow (9t-1)(9t-1)=0$

$\Rightarrow 9t-1=0$

$\Rightarrow 9t=1$

$\therefore t=\frac{1}{9}$

Now, from equation $\left(i\right)$,

$3^x=\frac{1}{9}$

$3^x=3^{-2}$

On comparing both sides, we get

$x=-2$