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If $3^{x+1} = 9^{x-2}$, find the value of $2^{1 +x}$.
Given:
$3^{x+1} = 9^{x-2}$.
To do:
We have to find the value of $2^{1 +x}$.
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,
$3^{x+1}=9^{x-2}$
$\Rightarrow 3^{x+1}=(3^{2})^{x-2}$
$\Rightarrow 3^{x+1}=3^{2 x-4}$
Comparing both sides, we get,
$x+1=2x-4$
$\Rightarrow 2 x-x=1+4$
$\Rightarrow x=5$
Therefore,
$2^{1+x}=2^{1+5}$
$=2^{6}$
$=64$
The value of $2^{1+x}$ is $64$.
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