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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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