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Solving Exponential Equations

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Algebra 2
Exponential and Logarithmic Functions
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Solving Exponential Equations 1.8 - Solution

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We have been given two equivalent expressions with the same base. Since both sides of the exponential equation are equal, the exponents must also be equal. 73x+5=71−x⇔3x+5=1−x\begin{gathered} 7^{3x+5} = 7^{1-x} \quad \Leftrightarrow \quad 3x+5=1-x \end{gathered} We can now solve the above equation.
3x+5=1−x3x+5=1-x
AddEqnLHS+x=RHS+x\text{LHS}+x=\text{RHS}+x
4x+5=14x+5=1
SubEqnLHS−5=RHS−5\text{LHS}-5=\text{RHS}-5
4x=-44x=\text{-}4
DivEqnLHS/4=RHS/4\left.\text{LHS}\middle/4\right.=\left.\text{RHS}\middle/4\right.
x=-1x=\text{-}1