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Solve the following equations for $x$:
\( 2^{x+1}=4^{x-3} \)
Given:
\( 2^{x+1}=4^{x-3} \)
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,
$2^{x+1}=4^{x-3}$
$2^{x+1}=(2^{2})^{x-3}$
$2^{x+1}=2^{2 x-6}$
Comparing the powers on both sides, we get,
$x+1=2 x-6$
$2 x-x=1+6$
$x=7$
Therefore, the value of $x$ is $7$.
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