4.6: Exponential and Logarithmic Equations
Using Like Bases to Solve Exponential Equations
The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers \(b\), \(S\), and \(T\), where \(b>0\), \(b≠1\), \(b^S=b^T\) if and only if \(S=T\).
In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.
For example, consider the equation \(3^{4x−7}=\dfrac{3^{2x}}{3}\). To solve for \(x\), we use the division property of exponents to rewrite the right side so that both sides have the common base, \(3\). Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for \(x\):
\[\begin{align*} 3^{4x-7}&= \dfrac{3^{2x}}{3}\\ 3^{4x-7}&= \dfrac{3^{2x}}{3^1} \qquad &&\text{Rewrite 3 as } 3^1\\ 3^{4x-7}&= 3^{2x-1} \qquad &&\text{Use the division property of exponents}\\ 4x-7&= 2x-1 \qquad &&\text{Apply the one-to-one property of exponents}\\ 2x&= 6 \qquad &&\text{Subtract 2x and add 7 to both sides}\\ x&= 3 \qquad &&\text{Divide by 3} \end{align*}\]
For any algebraic expressions \(S\) and \(T\), and any positive real number \(b≠1\),
\[\begin{align} b^S=b^T\text{ if and only if } S=T \end{align}\]
- Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form \(b^S=b^T\).
- Use the one-to-one property to set the exponents equal.
- Solve the resulting equation, \(S=T\), for the unknown.
Solve \(2^{x−1}=2^{2x−4}\).
Solution
\[\begin{align*} 2^{x-1}&= 2^{2x-4} \qquad &&\text{The common base is 2}\\ x-1&= 2x-4 \qquad &&\text{By the one-to-one property the exponents must be equal}\\ x&= 3 \qquad &&\text{Solve for x} \end{align*}\]
Solving Exponential Equations Using Logarithms
Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since \(\log(a)=\log(b)\) is equivalent to \(a=b\), we may apply logarithms with the same base on both sides of an exponential equation.
-
Apply the logarithm of both sides of the equation.
- If one of the terms in the equation has base 10, use the common logarithm.
- If none of the terms in the equation has base 10, use the natural logarithm.
- Use the rules of logarithms to solve for the unknown.
Solve \(5^{x+2}=4^x\).
Solution
\[\begin{align*}
5^{x+2}&= 4^x \qquad &&\text{There is no easy way to get the powers to have the same base}\\
\ln5^{x+2}&= \ln4^x \qquad &&\text{Take ln of both sides}\\
(x+2)\ln5&= x\ln4 \qquad &&\text{Use laws of logs}\\
x\ln5+2\ln5&= x\ln4 \qquad &&\text{Use the distributive law}\\
x\ln5-x\ln4&= -2\ln5 \qquad &&\text{Get terms containing x on one side, terms without x on the other}\\
x(\ln5-\ln4)&= -2\ln5 \qquad &&\text{On the left hand side, factor out an x}\\
x\ln \left (\dfrac{5}{4} \right )&= \ln \left (\dfrac{1}{25} \right ) \qquad &&\text{Use the laws of logs}\\
x&=\dfrac{\ln \left (\dfrac{1}{25} \right )}{\ln \left (\dfrac{5}{4} \right )} \qquad &&\text{Divide by the coefficient of x}
\end{align*}\]
Solve \(2^x=3^{x+1}\).
- Answer
-
\(x=\dfrac{\ln3}{\ln \left (\dfrac{2}{3} \right )}\)
Equations Containing \(e\)
One common type of exponential equations are those with base \(e\). This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base \(e\) on either side, we can use the natural logarithm to solve it.
- Divide both sides of the equation by \(A\).
- Apply the natural logarithm of both sides of the equation.
- Divide both sides of the equation by \(k\).
Solve \(100=20e^{2t}\).
Solution
\[\begin{align*} 100&= 20e^{2t}\\ 5&= e^{2t} \qquad &&\text{Divide by the coefficient of the power}\\ \ln5&= 2t \qquad &&\text{Take ln of both sides. Use the fact that } ln(x) \text{ and } e^x \text{ are inverse functions}\\ t&= \dfrac{\ln5}{2} \qquad &&\text{Divide by the coefficient of t} \end{align*}\]
Analysis
Using laws of logs, we can also write this answer in the form \(t=\ln\sqrt{5}\). If we want a decimal approximation of the answer, we use a calculator.
Using the Definition of a Logarithm to Solve Logarithmic Equations
We have already seen that every logarithmic equation \({\log}_b(x)=y\) is equivalent to the exponential equation \(b^y=x\). We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.
For example, consider the equation \({\log}_2(2)+{\log}_2(3x−5)=3\). To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for \(x\):
\[\begin{align*} {\log}_2(2)+{\log}_2(3x-5)&= 3\\ {\log}_2(2(3x-5))&= 3 \qquad \text{Apply the product rule of logarithms}\\ {\log}_2(6x-10)&= 3 \qquad \text{Distribute}\\ 2^3&= 6x-10 \qquad \text{Apply the definition of a logarithm}\\ 8&= 6x-10 \qquad \text{Calculate } 2^3\\ 18&= 6x \qquad \text{Add 10 to both sides}\\ x&= 3 \qquad \text{Divide by 6} \end{align*}\]
For any algebraic expression \(S\) and real numbers \(b\) and \(c\),where \(b>0\), \(b≠1\),
\[\begin{align} {\log}_b(S)=c \text{ if and only if } b^c=S \end{align}\]
Solve \(2\ln x+3=7\).
Solution
\[\begin{align*} 2\ln x+3&= 7\\ 2\ln x&= 4 \qquad \text{Subtract 3}\\ \ln x&= 2 \qquad \text{Divide by 2}\\ x&= e^2 \qquad \text{Rewrite in exponential form} \end{align*}\]
Using the One-to-One Property of Logarithms to Solve Logarithmic Equations
As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers \(x>0\), \(S>0\), \(T>0\) and any positive real number \(b\), where \(b≠1\),
\({\log}_bS={\log}_bT\) if and only if \(S=T\).
For example,
If \({\log}_2(x−1)={\log}_2(8)\), then \(x−1=8\).
So, if \(x−1=8\), then we can solve for \(x\),and we get \(x=9\). To check, we can substitute \(x=9\) into the original equation: \({\log}_2(9−1)={\log}_2(8)=3\). In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.
For example, consider the equation \(\log(3x−2)−\log(2)=\log(x+4)\). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for \(x\):
\[\begin{align*} \log(3x-2)-\log(2)&= \log(x+4)\\ \log \left (\dfrac{3x-2}{2} \right )&= \log(x+4) \qquad \text{Apply the quotient rule of logarithms}\\ \dfrac{3x-2}{2}&= x+4 \qquad \text{Apply the one to one property of a logarithm}\\ 3x-2&= 2x+8 \qquad \text{Multiply both sides of the equation by 2}\\ x&= 10 \qquad \text{Subtract 2x and add 2} \end{align*}\]
To check the result, substitute \(x=10\) into \(\log(3x−2)−\log(2)=\log(x+4)\).
\[\begin{align*} \log(3(10)-2)-\log(2)&= \log((10)+4) \\ \log(28)-\log(2)&= \log(14)\\ \log \left (\dfrac{28}{2} \right )&= \log(14) \qquad \text{The solution checks} \end{align*}\]
For any algebraic expressions \(S\) and \(T\) and any positive real number \(b\), where \(b≠1\),
\[\begin{align} b^S=b^T\text{ if and only if } S=T \end{align}\]
Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.
- Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form \({\log}_bS={\log}_bT\).
- Use the one-to-one property to set the arguments equal.
- Solve the resulting equation, \(S=T\), for the unknown.
Solve \(\ln(x^2)=\ln(2x+3)\).
Solution
\[\begin{align*} \ln(x^2)&= \ln(2x+3)\\ x^2&= 2x+3 \qquad \text{Use the one-to-one property of the logarithm}\\ x^2-2x-3&= 0 \qquad \text{Get zero on one side before factoring}\\ (x-3)(x+1)&= 0 \qquad \text{Factor using FOIL}\\ x-3&= 0 \qquad \text{or } x+1=0 \text{ If a product is zero, one of the factors must be zero}\\ x=3 \qquad \text{or} \\ x&= -11 \qquad \text{Solve for x} \end{align*}\]
Analysis
There are two solutions: \(3\) or \(−1\). The solution \(−1\) is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.
Solve \(\ln(x^2)=\ln1\).
- Answer
-
\(x=1\) or \(x=−1\)
Solving Applied Problems Using Exponential and Logarithmic Equations
In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.
One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . Table \(\PageIndex{1}\) lists the half-life for several of the more common radioactive substances.
| Substance | Use | Half-life |
|---|---|---|
| gallium-67 | nuclear medicine | 80 hours |
| cobalt-60 | manufacturing | 5.3 years |
| technetium-99m | nuclear medicine | 6 hours |
| americium-241 | construction | 432 years |
| carbon-14 | archeological dating | 5,715 years |
| uranium-235 | atomic power | 703,800,000 years |
We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:
\[\begin{align} A(t)&= A_0e^{\tfrac{\ln(0.5)}{T}t}\\ A(t)&= A_0e^{\tfrac{\ln(0.5)t}{T}}\\ A(t)&= A_0{(e^{\ln(0.5)})}^{\tfrac{t}{T}}\\ A(t)&= A_0{\left (\dfrac{1}{2}\right )}^{\tfrac{t}{T}}\\ \end{align}\]
where
- \(A_0\) is the amount initially present
- \(T\) is the half-life of the substance
- \(t\) is the time period over which the substance is studied
- \(y\) is the amount of the substance present after time \(t\)
How long will it take for ten percent of a \(1000\)-gram sample of uranium-235 to decay?
Solution
\[\begin{align*} y&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t}\\ 900&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{After } 10\% \text{ decays, 900 grams are left}\\ 0.9&= e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{Divide by 1000}\\ \ln(0.9)&= \ln \left (e^{\tfrac{\ln(0.5)}{703,800,000}t} \right ) \qquad \text{Take ln of both sides}\\ \ln(0.9)&= \dfrac{\ln(0.5)}{703,800,000}t \qquad \ln(e^M)=M\\ t&= 703,800,000\times \dfrac{\ln(0.9)}{\ln(0.5)} \qquad \text{years Solve for t}\\ t&\approx 106,979,777 \qquad \text{years} \end {align*} \]
Analysis
Ten percent of \(1000\) grams is \(100\) grams. If \(100\) grams decay, the amount of uranium-235 remaining is \(900\) grams.