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Solve the following equations for $x$:
\( 4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{x} \)
Given:
\( 4^{x-1} \times(0.5)^{3-2 x}=\left(\frac{1}{8}\right)^{x} \)
To do:
We have to find the value of $x$.
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,
$4^{x-1} \times(0.5)^{3-2 x}=(\frac{1}{8})^{x}$
$(2^2)^{x-1}\times(\frac{1}{2})^{3-2x}=(\frac{1}{2^3})^{x}$
$(2)^{2x-2}\times(2^{-1})^{3-2x}=(2^{-3})^{x}$
$(2)^{2x-2}\times(2)^{-3+2x}=(2)^{-3x}$
$(2)^{2x-2-3+2x}=(2)^{-3x}$
$(2)^{4x-5}=(2)^{-3x}$
Comparing the powers on both sides, we get,
$4x-5=-3x$
$4x+3x=5$
$7x=5$
$x=\frac{5}{7}$
Therefore, the value of $x$ is $\frac{5}{7}$.
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