Determine $(8x)^x$, if $9^{x+ 2} = 240 + 9^x$.

Given:

$9^{x+ 2} = 240 + 9^x$.

To do: 

We have to find the value of $(8x)^x$.

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,

$9^{x+2}=240+9^{x}$

$9^{x} \times 9^{2}=240+9^{x}$

$\Rightarrow 9^{x} \times 81=9^{x}+240$

$9^{x}\times81-9^{x}=240$

$\Rightarrow 9^{x}(81-1)=240$

$\Rightarrow 9^{x}=\frac{240}{80}$

$\Rightarrow 9^{x}=3$

$\Rightarrow 3^{2 x}=3^{1}$

Comparing both sides, we get,

$2 x=1$

$\Rightarrow x=\frac{1}{2}$

Therefore,

$(8 x)^{x}=(8 \times \frac{1}{2})^{\frac{1}{2}}$

$=4^{\frac{1}{2}}$

$=(2^{2})^{\frac{1}{2}}$

$=2^{1}$

$=2$

The value of $(8x)^x$ is $2$.

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Updated on: 10-Oct-2022

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