Question

Solve the following equations for x:

(i) $2^{2x}-2^{x+3}+2^4$ =0

(ii) $3^{2x+4}+1=2.3^{x+2}$

Answer 1

(i) $2^{2x}-2^{x+3}+2^4=0$

$\Rightarrow (2^x{)}^2-2^3\times (2{)}^x+2^4=0$

$\Rightarrow (2^x{)}^2-8(2{)}^x+16=0$

$\Rightarrow$ $(2^x{)}^2-2\times 2^x\times 4+(4{)}^2=0$

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

$\Rightarrow 2^x-4=0$

$\Rightarrow 2^x=4=2^2$

Comparing we get, $x=2$

(ii) $3^{2x+4}+1=2.3^{x+2}$

$\Rightarrow 3^{2x}\times 3^4+1-2.3^x\times 3^2=0$

$\Rightarrow 81.3^{2x}-2\times 9\left(3^x\right)+1=0$

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

$\Rightarrow \left[9\right(3^x){]}^2-2\times 9\times 3^x+(1{)}^2=0$

$\Rightarrow 9(3{)}^x-1=0\Rightarrow 9(3^x)=1$

$\Rightarrow 3^x=\frac{1}{9}=\frac{1}{3^2}=3^{-2}$
Comparing,we get
x=-2