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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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