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College Algebra/Solving Exponential and Logarithmic Equations/Exponential Equations

Solving Exponential and Logarithmic Equations

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Exponential Equations

Solving Exponential Equations Algebraically

Exponential equations can be solved by using equality properties of exponential expressions or by using the relationship between exponents and logarithms.
An exponential equation is an equation with an exponential expression that has a variable in the exponent. An exponential equation may have exponential expressions on one side, such as:
3x=73^x=7
It can also have the exponential expressions on both sides, such as:
6x+2=42x+36^{x+2}=4^{2x+3}
Solving an exponential equation with exponential expressions on both sides may involve the equality property: If bx=byb^x=b^y, then x=yx=y.
Step-By-Step Example
Solve Exponential Equations Using the Equality Property
Solve for xx.
22x+8=43x2^{2x+8}=4^{3x}
Step 1

Write the exponential expressions with the same base.

The bases are 2 and 4. Since 4 can be written as 222^2, substitute 222^2 for 4.
22x+8=43x22x+8=(22)3x22x+8=22⋅3x22x+8=26x\begin{aligned}2^{2x+8}&=4^{3x}\\2^{2x+8}&=\left(2^2\right)^{3x}\\2^{2x+8}&=2^{2\cdot3x}\\2^{2x+8}&=2^{6x}\end{aligned}
Step 2

Use the equality property.

The expressions have the same base, so the exponents are equal.
2x+8=6x2x+8=6x
Solution
Solve the resulting equation.
2x+8=6x8=4x2=x\begin{aligned}2x+8&=6x\\8&=4x\\2&=x\end{aligned}

To solve an exponential equation with exponential expressions on one side, use the relationship between exponents and logarithms: If by=xb^y=x, then logbx=y{\rm{log}}_{b}\, x=y.
Step-By-Step Example
Solve Exponential Equations Using Logarithms
Solve for xx.
7x−2=127^{x-2}=12
Step 1
Write the equation in logarithmic form.
log⁡712=x−2\log_7 {12}=x-2
Step 2
Solve the equation for xx.
log⁡7(12)+2=x\log_7 {(12)}+2=x
Solution

Use technology to evaluate the logarithm, and simplify the expression for xx.

If needed, use the change of base rule to rewrite the logarithm with a base that a calculator can compute, such as base 10 or base ee.
log⁡(12)log⁡(7)+2=x1.28+2≈x3.28≈x\begin{aligned}\frac{\log{(12)}}{\log{(7)}}+2&=x\\1.28 + 2&\approx x\\3.28 &\approx x\end{aligned}
Step-By-Step Example
Solve Exponential Equations with Natural Logarithms
Solve for xx:
5ex=205e^{x}=20
Step 1

Isolate the exponential expression.

Divide both sides of the equation by 5.
ex=4e^{x}=4
Step 2

Write the equation in logarithmic form.

The base of the exponential expression is ee, so use the natural logarithm.
ln⁡4=x\ln {4}=x
Solution
Use technology to evaluate the logarithm.
1.39≈x1.39\approx x

Solving some exponential equations with exponential expressions on both sides requires taking the logarithm of both sides and then using properties of logarithms. When the logarithm is taken of both sides of the equation, the same base must be used. It is helpful to use a base that a calculator can easily compute, so choose log⁡\log or ln⁡\ln.
Step-By-Step Example
Solve Exponential Equations by Taking the Logarithm of Both Sides
Solve the equation for xx.
6x+2=42x+36^{x+2}=4^{2x+3}
Step 1
Take the logarithm of both sides of the equation.
log⁡6x+2=log⁡42x+3\log6^{x+2}=\log4^{2x+3}
Step 2
The power rule for logarithms is:
log⁡b(xp)=p⋅log⁡b(x)\log_{b}{(x^p)}=p\cdot\log_{b}{(x)}
Use the properties of logarithms to write the expressions without exponents.
(x+2)log⁡6=(2x+3)log⁡4(x+2)\log6=(2x+3)\log4
Step 3
Use the distributive property.
xlog⁡6+2log⁡6=2xlog⁡4+3log⁡4x\log6 + 2\log6 = 2x\log4 + 3\log4
Step 4
Rewrite the equation with variable terms on one side.
2log⁡6−3log⁡4=2xlog⁡4−xlog⁡62\log6-3\log4 = 2x\log4 -x\log6
Step 5
Isolate the variable.
2log⁡6−3log⁡4=x(2log⁡4−log⁡6)Factor out x.2log⁡6−3log⁡42log⁡4−log⁡6=xDivide.\begin{aligned}2\log6-3\log4 &= x(2\log4 -\log6)\;\;\;\;\;&&\text{Factor out }\;x\text{.}\\\\{\frac{2\log6-3\log4}{2\log4 -\log6}} &= x\;\;\;\;\;&&\text{Divide.}\end{aligned}
Solution
Use technology to evaluate the logarithms and find an approximate solution.
−0.587≈x-0.587\approx x

Exponential equations are used in a wide variety of scientific fields to model data and make predictions about the natural world. For example, radiocarbon dating is based on the application of an exponential model.

Radiocarbon dating is a method of approximating the age of an object that contains organic material. Carbon-14 is a radioactive isotope of carbon that decays into another element over time. By measuring the amount of carbon-14 left in an artifact and comparing that to the amount that would be expected in a living organism, scientists can determine how much carbon-14 has decayed and then calculate how much time it would have taken for that amount of carbon-14 to decay. This gives an estimate of how old the object is.

Step-By-Step Example
Solve an Exponential Equation Modeling Decay
An archaeologist discovers a fossil that contains 25% of the amount of carbon-14 expected in a living organism. In the equation A=A0e−0.000124tA=A_{0}e^{-0.000124t}, A0A_{0} is the amount of carbon-14 expected in a living organism and AA is the amount of carbon-14 remaining in the object after tt years. Use the equation to approximate the age of the fossil.
Step 1
The fossil contains 25% of the amount of carbon-14 expected in a living organism, so the ratio of the amount AA in the fossil compared to the amount A0A_0 expected in a living organism is 0.25. Write the equation in terms of this ratio.
A=A0e−0.000124tAA0=e−0.000124t\begin{aligned}A&=A_{0}e^{-0.000124t}\\\frac{A}{A_{0}}&=e^{-0.000124t}\end{aligned}
Substitute 0.25 for the ratio.
0.25=e−0.000124t0.25=e^{-0.000124t}
Step 2

Take the logarithm of both sides of the equation. Then simplify.

The base of the exponential expression is ee, so use the natural logarithm.
ln⁡0.25=ln⁡e−0.000124tln⁡0.25=−0.000124t\begin{aligned}\ln{0.25}&=\ln{e^{-0.000124t}}\\\ln{0.25}&=-0.000124t\end{aligned}
Step 3
Use technology to evaluate the logarithm.
ln⁡0.25≈−1.39\ln{0.25}\approx -1.39
Substitute –1.39 for the logarithm.
−1.39=−0.000124t-1.39=-0.000124t
Solution
Solve the equation for tt by dividing both sides by –0.000124.
11,210≈t11\rm{,}210\approx t
The fossil is approximately 11,210 years old.

Solving Exponential Equations Using Substitution

Some exponential equations can be solved by using substitution to change the form of the equation.
Some exponential equations can be solved by replacing an exponential expression with a temporary variable. Solve for the temporary variable, and then replace the temporary variable with the exponential expression and solve for xx.
Step-By-Step Example
Solve Exponential Equations Using Substitution
Solve the equation for xx:
e2x−5ex−24=0e^{2x}-5e^x-24 = 0
Step 1

Use a temporary variable to represent an exponential expression.

Substitute uu for exe^x.
e2x−5ex−24=0u2−5u−24=0\begin{aligned}e^{2x}-5e^x-24 &= 0\\u^2-5u-24 &= 0\end{aligned}
Step 2
Factor the equation.
u2−5u−24=0(u−8)(u+3)=0\begin{aligned} u^2-5u-24 &= 0\\(u - 8)(u + 3) &= 0\end{aligned}
Step 3
Use the zero product property to set each factor equal to zero. Then, solve each equation.
u−8=0u=8oru+3=0u=−3\begin{aligned}u-8&=0\\u&=8\end{aligned}\hspace{10pt}\text{or}\hspace{10pt}\begin{aligned} u+3&=0\\u&=-3\end{aligned}
Step 4
Replace uu with the original expression it represents, exe^x.
u=8ex=8oru=−3ex=−3\begin{aligned}u&=8\\e^x &= 8 \end{aligned} \hspace{10pt} \text{or} \hspace{10pt}\begin{aligned}u&=-3\\e^x &= -3\end{aligned}
Step 5
Write the equations in logarithmic form.
ex=8x=ln⁡8orex=−3x=ln⁡(−3)\begin{aligned}e^x &= 8\\x&=\ln8\end{aligned} \hspace{10pt}\text{or} \hspace{10pt} \begin{aligned}e^x &= -3\\x&=\ln{(-3)}\end{aligned}
All arguments of logarithms must be positive, so x=ln⁡(−3)x=\ln{(-3)} is not a valid solution.
Solution
Use technology to evaluate the logarithm in the valid solution.
x≈2.079x\approx2.079

Solving Exponential Equations by Graphing

Exponential equations can be solved by graphing the related functions for both sides of the equation and looking for points of intersection.
To solve an exponential equation by graphing, graph the related functions for both sides of the equation and look for points of intersection, as when solving a system of equations in two variables. If the exponential equation is set equal to zero, graph the related function for the exponential expression and look for the zeros.
Step-By-Step Example
Solve Exponential Equations by Graphing
Solve the equation for xx.
42x+3=24^{2x+3}=2
Step 1
Graph the functions by using a graphing calculator or other graphing utility:
f(x)=42x+3f(x)=2f(x)=4^{2x+3} \hspace{20pt} f(x)=2
Step 2
Use the trace function on a graphing utility to approximate the xx-value of the point of intersection.
Solution
The graphs intersect at the point (−1.25,2)(-1.25,2). So the solution is:
x=−1.25x=-1.25
<Logarithmic Equations