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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Updated on: 10-Oct-2022

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