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Solving Exponential Equations

Solving Exponential Equations 1.9 - Solution

arrow_back Return to Solving Exponential Equations
To solve the given exponential equation, we will start by rewriting the terms so that they have a common base.
(15)x5=253x+2\left(\dfrac{1}{5}\right)^{x-5}=25^{3x+2}
(151)x5=(52)3x+2\left(\dfrac{1}{5^1}\right)^{x-5}=\left(5^2\right)^{3x+2}
(5-1)x5=(52)3x+2\left(5^{\text{-} 1}\right)^{x-5}=\left(5^2\right)^{3x+2}
5-1(x5)=52(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. 5-1(x5)=52(3x+2)-1(x5)=2(3x+2)\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 -1(x5)=2(3x+2).\text{-} 1(x-5)=2(3x+2).
-1(x5)=2(3x+2)\text{-} 1(x-5)=2(3x+2)
Solve for xx
-x+5=2(3x+2)\text{-} x+5=2(3x+2)
-x+5=6x+4\text{-} x+5=6x+4
-x=6x1\text{-} x=6x-1
-7x=-1\text{-} 7x=\text{-} 1
x=-1-7x=\dfrac{\text{-} 1}{\text{-} 7}
x=17x=\dfrac{1}{7}