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Solving Exponential Equations
Choose Course
Algebra 2
Exponential and Logarithmic Functions
Solving Exponential Equations
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Solving Exponential Equations 1.8 - Solution
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Return to Solving Exponential Equations
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.
7
3
x
+
5
=
7
1
−
x
⇔
3
x
+
5
=
1
−
x
\begin{gathered} 7^{3x+5} = 7^{1-x} \quad \Leftrightarrow \quad 3x+5=1-x \end{gathered}
7
3
x
+
5
=
7
1
−
x
⇔
3
x
+
5
=
1
−
x
We can now solve the above equation.
3
x
+
5
=
1
−
x
3x+5=1-x
3
x
+
5
=
1
−
x
AddEqn
LHS
+
x
=
RHS
+
x
\text{LHS}+x=\text{RHS}+x
LHS
+
x
=
RHS
+
x
4
x
+
5
=
1
4x+5=1
4
x
+
5
=
1
SubEqn
LHS
−
5
=
RHS
−
5
\text{LHS}-5=\text{RHS}-5
LHS
−
5
=
RHS
−
5
4
x
=
-
4
4x=\text{-}4
4
x
=
-
4
DivEqn
LHS
/
4
=
RHS
/
4
\left.\text{LHS}\middle/4\right.=\left.\text{RHS}\middle/4\right.
LHS
/
4
=
RHS
/
4
x
=
-
1
x=\text{-}1
x
=
-
1