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Solve the following equations for $x$:$ 7^{2 x+3}=1 $
Given:
\( 7^{2 x+3}=1 \)
To do:
We have to find the value of $x$.
Solution:
We know that,
$(a^{m})^{n}=a^{m n}$
$a^{m} \times a^{n}=a^{m+n}$
$a^{m} \div a^{n}=a^{m-n}$
$a^{0}=1$
$7^{2 x+3}=1$
$7^{2 x+3}=7^{0}$
Comparing the powers on both sides, we get,
$2 x+3=0$
$2 x=-3$
$x=\frac{-3}{2}$
Therefore, the value of $x$ is $\frac{-3}{2}$.
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