Answer
$x=\dfrac{1}{2}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\left( \dfrac{1}{8} \right)^{2x-3}=16^{x+1}
,$ express both sides in the same base. Once the bases are the same, equate the exponents. Use the properties of equality to isolate the variable.
$\bf{\text{Solution Details:}}$
Using exponents, the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{1}{2^3} \right)^{2x-3}=(2^4)^{x+1}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
\left( 2^{-3} \right)^{2x-3}=(2^4)^{x+1}
.\end{array}
Using the Power Rule of the laws of exponents which is given by $\left( x^m \right)^p=x^{mp},$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
2^{-3(2x-3)}=2^{4(x+1)}
\\\\
2^{-6x+9}=2^{4x+4}
.\end{array}
Since the bases are the same, then the exponents can be equated. Hence,
\begin{array}{l}\require{cancel}
-6x+9=4x+4
\\\\
-6x-4x=4-9
\\\\
-10x=-5
\\\\
x=\dfrac{-5}{-10}
\\\\
x=\dfrac{1}{2}
.\end{array}
