Solving Exponential Equations

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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.

Solving Exponential Equations Graphically

If the dependent variable of an exponential function written in the form y=abx, y = a \cdot b^x, is exchanged for a constant, say C,C, the result is a one-variable equation: C=abx. 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=abx,y = a \cdot b^x, then finding the xx-coordinate of the point(s) on the graph with the yy-coordinate C.C. The xx-coordinate(s) is the solution to the equation.

Use the graph to solve the equation 3=50.85x.3 = 5 \cdot 0.85^x.


The graph shows all xx-yy points that satisfy the function rule y=50.85x.y = 5 \cdot 0.85^x. Let's compare the function rule and the equation. Function rule:y=50.85xEquation:3=50.85x\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,y, is replaced by a 33 in the equation. Thus, we solve the equation by finding the xx-coordinate of any point on the graph that has the yy-coordinate 3.3.

We can identify one such point in the graph. Let's now find the xx-coordinate of this point graphically.

This xx-coordinate is not easily read from the graph, so we'll have to make an approximation. It's just a bit bigger than 3,3, so we'll use This means that an approximate solution to the equation is x3.1.x \approx 3.1. We can verify this by substituting it into equation to see if a true statement is made.

3=50.85x3 = 5 \cdot 0.85^x
x3.1x \approx {\color{#0000FF}{3.1}}
3?50.853.13 \overset{?}{\approx} 5 \cdot 0.85^{{\color{#0000FF}{3.1}}}
33.021123 \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: x3.1.x \approx 3.1.

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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 42x=642. 4^{2x}=64^2. Since 64=43,64=4^3, the equation can be written as follows. 42x=(43)242x=46\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=6x=3. 2x=6 \quad \Leftrightarrow \quad x=3.

Solve the equation 35x3=3-2x+11. 3^{5x-3}=3^{\text{-}2x+11}.

To begin, notice that both sides of the equation are exponential expressions with base 3.3. Since they have the same base, the exponents must be equal. This gives the equivalent equation 5x3=-2x+11, 5x-3=\text{-}2x+11, which we can solve using inverse operations.
The solution to the equation is x=2.x=2.
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Solving an Exponential Equation with Logarithms

When the expressions on both sides of an exponential equation cannot be written with the same base, logarithms can used instead. Essentially, a logarithm is used to undo the exponent. Consider the equation 8x=3. 8^x=3. Since xx is the exponent of base 8,8, log8\log_8 can be applied to each side of the equation. This yields log8(8x)=log8(3). {\color{#0000FF}{\log_8}}\left(8^x\right)={\color{#0000FF}{\log_8}}(3). Now, log8(8x)\log_8\left(8^x\right) is equal to the exponent needed to make 88 equal 8x.8^x. This can also be expressed as log8(8x)=x.\log_8\left(8^x\right)=x. By using this to rewrite the left-hand side the equation becomes x=log8(3). x=\log_8(3). This is the exact solution to the equation. If a numerical value is required, use a calculator. However, the "log" button on a calculator corresponds to a common logarithm — log10.\log_{10}. When the logarithm has a base different than 10,10, the Change of Base rule can be used to express the logarithm with base 10.10. logn(m)=log(m)log(n). \log_n(m)=\dfrac{\log(m)}{\log(n)}. For the example, xx equals

log8(3)=log10(3)log10(8)0.53. \log_8(3)=\dfrac{\log_{10}(3)}{\log_{10}(8)}\approx 0.53.


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