Solving Exponential Equations

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}

WritePowWrite as a power

\left(\dfrac{1}{5^1}\right)^{x-5}=\left(5^2\right)^{3x+2}

\dfrac{1}{a^m}=a^{\text{-} m}

\left(5^{\text{-} 1}\right)^{x-5}=\left(5^2\right)^{3x+2}

\left(a^{m}\right)^{n}=a^{m\cdot n}

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

DistrDistribute

\text{-} 1

\text{-} x+5=2(3x+2)

DistrDistribute

2

\text{-} x+5=6x+4

\text{LHS}-5=\text{RHS}-5

\text{-} x=6x-1

\text{LHS}-6x=\text{RHS}-6x

\text{-} 7x=\text{-} 1

\left.\text{LHS}\middle/(\text{-} 7)\right.=\left.\text{RHS}\middle/(\text{-} 7)\right.

x=\dfrac{\text{-} 1}{\text{-} 7}

\dfrac{\text{-} a}{\text{-} b}=\dfrac{a}{b}

x=\dfrac{1}{7}

Solving Exponential Equations 1.9 - Solution