Solve the following equations for $x$:$ 3^{2 x+4}+ 1=2 \cdot 3^{x+2} $

Given:

\( 3^{2 x+4}+ 1=2 \cdot 3^{x+2} \)

To do:

We have to find the value of $x$.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$  

Therefore,

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

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

$\Rightarrow 81(3^{2 x})-2 \times 9(3^{x})+1=0$

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

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

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

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

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

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

$\Rightarrow 3^{x}=\frac{1}{3^{2}}$

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

Comparing the powers on both sides, we get,

$x=-2$

Therefore, the value of $x$ is $-2$.   

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Updated on: 10-Oct-2022

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