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{{ courseTrack.displayTitle }} {{ printedBook.courseTrack.name }} {{ printedBook.name }} # Solving Exponential Equations

Exponential equations — equations in which the independent variable is an exponent — can be solved graphically and algebraically. Depending on the equation, different algebraic approaches can be used.
Method

## Solving Exponential Equations Graphically

If the dependent variable of an exponential function written in the form $y = a \cdot b^x,$ is exchanged for a constant, say $C,$ the result is a one-variable equation: $C = a \cdot b^x.$

This type of equation is called an exponential equation, and can be solved graphically. This is done by first graphing the function $y = a \cdot b^x,$ then finding the $x$-coordinate of the point(s) on the graph with the $y$-coordinate $C.$ The $x$-coordinate(s) is the solution to the equation.
Exercise

Use the graph to solve the equation $3 = 5 \cdot 0.85^x.$ Solution

The graph shows all $x$-$y$ points that satisfy the function rule $y = 5 \cdot 0.85^x.$ Let's compare the function rule and the equation. \begin{aligned} \textbf{Function rule:} \quad y = 5 \cdot 0.85^x\\ \textbf{Equation:} \quad 3 = 5 \cdot 0.85^x \end{aligned} The only difference between these two equalities is that the independent variable, $y,$ is replaced by a $3$ in the equation. Thus, we solve the equation by finding the $x$-coordinate of any point on the graph that has the $y$-coordinate $3.$ We can identify one such point in the graph. Let's now find the $x$-coordinate of this point graphically. This $x$-coordinate is not easily read from the graph, so we'll have to make an approximation. It's just a bit bigger than $3,$ so we'll use $3.1.$ This means that an approximate solution to the equation is $x \approx 3.1.$ We can verify this by substituting it into equation to see if a true statement is made.

$3 = 5 \cdot 0.85^x$
$x \approx {\color{#0000FF}{3.1}}$
$3 \overset{?}{\approx} 5 \cdot 0.85^{{\color{#0000FF}{3.1}}}$
$3 \approx 3.02112 \ldots$

The right-hand side and the left-hand side are approximately equal, so we have indeed found an approximate solution to the equation: $x \approx 3.1.$

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Method

## Solving Exponential Equations with the Same Base

There are different ways to algebraically solve exponential equations. If both sides of the equation can be written in the same base, equality can be used. For example, consider the equation $4^{2x}=64^2.$ Since $64=4^3,$ the equation can be written as follows. \begin{aligned} 4^{2x} &=\left({\color{#0000FF}{4^3}} \right)^{2}\\ 4^{2x} &=4^6\\ \end{aligned} Now, we have two equivalent expressions with the same base. For the equality to hold, the exponents must also be equal. Thus,

$2x=6 \quad \Leftrightarrow \quad x=3.$
Exercise

Solve the equation $3^{5x-3}=3^{\text{-}2x+11}.$

Solution
To begin, notice that both sides of the equation are exponential expressions with base $3.$ Since they have the same base, the exponents must be equal. This gives the equivalent equation $5x-3=\text{-}2x+11,$ which we can solve using inverse operations.
$5x-3=\text{-}2x+11$
$7x-3=11$
$7x=14$
$x=2$
The solution to the equation is $x=2.$
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