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Solve the following equations:$ 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 solve the given equation.
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,
$4^{x-1} \times(0.5)^{3-2 x}=(\frac{1}{8})^{x}$
$\Rightarrow (2^{2})^{x-1} \times(\frac{1}{2})^{3-2 x}=(\frac{1}{2^{3}})^{x}$
$\Rightarrow 2^{2 x-2} \times 2^{-3+2 x}=2^{-3 x}$
$\Rightarrow 2^{2 x-2-3+2 x}=2^{-3 x}$
$\Rightarrow 2^{4 x-5}=2^{-3 x}$
Comparing both sides, we get,
$4 x-5=-3 x$
$\Rightarrow 4 x+3 x=5$
$\Rightarrow 7 x=5$
$\Rightarrow x=\frac{5}{7}$
The values of $x$ is $\frac{5}{7}$.
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