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# Solving Exponential Equations

## Solving Exponential Equations 1.9 - Solution

To solve the given exponential equation, we will start by rewriting the terms so that they have a common base.
$\left(\dfrac{1}{5}\right)^{x-5}=25^{3x+2}$
$\left(\dfrac{1}{5^1}\right)^{x-5}=\left(5^2\right)^{3x+2}$
$\left(5^{\text{-} 1}\right)^{x-5}=\left(5^2\right)^{3x+2}$
$5^{\text{-} 1(x-5)}=5^{2(3x+2)}$
Now, we have two equivalent expressions with the same base. If both sides of the equation are equal, the exponents must also be equal. $\begin{gathered} 5^{{\color{#0000FF}{\text{-} 1(x-5)}}}=5^{{\color{#0000FF}{2(3x+2)}}} \quad \Leftrightarrow \quad {\color{#0000FF}{\text{-} 1(x-5)}}={\color{#0000FF}{2(3x+2)}} \end{gathered}$ Finally, we will solve the equation $\text{-} 1(x-5)=2(3x+2).$
$\text{-} 1(x-5)=2(3x+2)$
Solve for $x$
$\text{-} x+5=2(3x+2)$
$\text{-} x+5=6x+4$
$\text{-} x=6x-1$
$\text{-} 7x=\text{-} 1$
$x=\dfrac{\text{-} 1}{\text{-} 7}$
$x=\dfrac{1}{7}$