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If $( \frac{1}{16})^{4-3x}\times 8^{x-2}=( 0.25)^x$, the find the value of $\frac{17x}{22}+1$.
Given: $( \frac{1}{16})^{4-3x}\times 8^{x-2}=( 0.25)^x$.
To do: To find the value of $\frac{17x}{22}+1$.
Solution:
$( \frac{1}{16})^{4-3x}\times 8^{x-2}=( 0.25)^x$
$\Rightarrow ( \frac{1}{2^4})^{4-3x}\times ( 2^3)^{x-2}=( \frac{25}{100})^x$
$\Rightarrow (2^{-4})^{4-3x}\times( 2^3)^{x-2}=( \frac{1}{4})^x$
$\Rightarrow ( 2)^{-4( 4-3x)}\times( 2)^{3( x-2)}=( \frac{1}{2^2})^x$
$\Rightarrow ( 2)^{( -16+12x)}\times( 2)^{( 3x-6)}=( 2)^{-2x}$
$\Rightarrow ( 2)^{( -16+12x+3x-6)}=( 2)^{-2x}$
$\Rightarrow ( 2)^{( 15x-22)}=( 2)^{-2x}$
$\Rightarrow 15x-22=-2x$
$\Rightarrow 15x+2x=22$
$\Rightarrow 17x=22$
$\Rightarrow x=\frac{22}{17}$
Therefore, $\frac{17x}{22}+1=\frac{17}{22}\times\frac{22}{17}+1$
$=1+1=2$
Thus, $\frac{17x}{22}+1=2$
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