Learning Objectives In 1859, an Australian landowner named Thomas Austin released \(24\) rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions. Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions. Solve Exponential EquationsUse the One-to-One Property of Exponential FunctionsThe 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 \(x\): \[\begin{align*} 3^{4x-7}&= \dfrac{3^{2x}}{3}\\ 3^{4x-7}&= \dfrac{3^{2x}}{3^1} \qquad &&\text{Rewrite 3 as } 3^1\\ 3^{4x-7}&= 3^{2x-1} \qquad &&\text{Use the division property of exponents}\\ 4x-7&= 2x-1 \qquad &&\text{Apply the one-to-one property of exponents}\\ 2x&= 6 \qquad &&\text{Subtract 2x and add 7 to both sides}\\ x&= 3 \qquad &&\text{Divide by 3} \end{align*}\] THE 1-1 PROPERTY OF EXPONENTIAL FUNCTIONS For any algebraic expressions \(S\) and \(T\), and any positive real number \(b≠1\), \[\begin{align} b^S=b^T\text{ if and only if } S=T \end{align}\]
Example \(\PageIndex{1}\): Solve an Exponential Equation with a Common Base Solve \(2^{x−1}=2^{2x−4}\). Solution \[\begin{align*} 2^{x-1}&= 2^{2x-4} \qquad &&\text{The common base is 2}\\ x-1&= 2x-4 \qquad &&\text{By the one-to-one property the exponents must be equal}\\ x&= 3 \qquad &&\text{Solve for x} \end{align*}\]
Solve \(5^{2x}=5^{3x+2}\). Answer\(x=−2\) Common Base MethodSometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property. For example, consider the equation \(256=4^{x−5}\). We can rewrite both sides of this equation as a power of \(2\). Then we apply the rules of exponents, along with the one-to-one property, to solve for \(x\): \[\begin{align*} 256&= 4^{x-5}\\ 2^8&= {(2^2)}^{x-5} \qquad &&\text{Rewrite each side as a power with base 2}\\ 2^8&= 2^{2x-10} \qquad &&\text{Use the one-to-one property of exponents}\\ 8&= 2x-10 \qquad &&\text{Apply the one-to-one property of exponents}\\ 18&= 2x \qquad &&\text{Add 10 to both sides}\\ x&= 9 \qquad &&\text{Divide by 2} \end{align*}\]
Example \(\PageIndex{2}\): Solve Equations by Rewriting Them to Have a Common Base Solve \(8^{x+2}={16}^{x+1}\). Solution \[\begin{align*} 8^{x+2}&= {16}^{x+1}\\ {(2^3)}^{x+2}&= {(2^4)}^{x+1} \qquad &&\text{Write 8 and 16 as powers of 2}\\ 2^{3x+6}&= 2^{4x+4} \qquad &&\text{To take a power of a power, multiply exponents}\\ 3x+6&= 4x+4 \qquad &&\text{Use the one-to-one property to set the exponents equal}\\ x&= 2 \qquad &&\text{Solve for } x \end{align*}\]
Solve \(5^{2x}={25}^{3x+2}\). Answer\(x=−1\) \(\PageIndex{3}\) Solve \(2^{5x}=\sqrt{2}\). Solution \[\begin{align*} 2^{5x}&= 2^{\frac{1}{2}} \qquad &&\text{Write the square root of 2 as a power of 2}\\ 5x&= \dfrac{1}{2} \qquad &&\text{Use the one-to-one property}\\ x&= \dfrac{1}{10} \qquad &&\text{Solve for } x \end{align*}\]
Solve \(5^x=\sqrt{5}\). Answer\(x=\dfrac{1}{2}\)
No. Recall that the range of an exponential function is always positive. In the process of solving an exponential equation, if the equation obtained is an exponential expression that is not equal to a positive number, there is no solution for that equation. Example \(\PageIndex{4}\): Exponential Equation with no solution Solve \(3^{x+1}=−2\). Solution This equation has no solution. There is no real value of \(x\) that will make the equation a true statement because any power of a positive number is positive. The figure below shows that the two graphs do not cross so the left side of the equation is never equal to the right side. Thus the equation has no solution.
Solve \(2^x=−100\). AnswerThe equation has no solution. Rewrite in Logarithmic FormSometimes 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.
Example \(\PageIndex{5}\): Solve an Equation Containing Powers of Different Bases Solve \(5^{x+2}=4^x\). Solution \[\begin{align*}
Solve \(2^x=3^{x+1}\). Answer\(x=\dfrac{\ln3}{\ln \left (\dfrac{2}{3} \right )}\)
Yes. The solution is \(x = 0\). (Take the \( \ln \) of both sides, use the power rule, and solve for \(x\)). 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.
Example \(\PageIndex{6}\): Solve an Equation of the Form \(y = Ae^{kt}\) 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 logarithm rules, this answer can be rewritten in the form \(t=\ln\sqrt{5}\). A calculator can be used to obtain a decimal approximation of the answer, \( t \approx 0.8047 \).
Solve \(3e^{0.5t}=11\). Answer\(t=2\ln \left (\dfrac{11}{3} \right )\) or \(\ln{ \left (\dfrac{11}{3} \right )}^2\)
No. When \(k≠0\), there is a solution when \(y\) and \(A\) are either both 0, or when neither is 0 and they have the same sign. An example of an equation with this form that does not have a solution is \(2=−3e^t\), which would mean \(e^t\) is negative, which is impossible. Example \(\PageIndex{7}\): Solve an Equation That Can Be Simplified to the Form \(y=Ae^{kt}\) Solve \(4e^{2x}+5=12\). Solution \[\begin{align*} 4e^{2x}+5&= 12\\ 4e^{2x}&= 7 \qquad &&\text{Combine like terms}\\ e^{2x}&= \dfrac{7}{4} \qquad &&\text{Divide by the coefficient of the power}\\ 2x&= \ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Take ln of both sides and use }\ln e^u = u\\ x&= \dfrac{1}{2}\ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Solve for x} \end{align*}\]
Solve \(3+e^{2t}=7e^{2t}\). Answer\(t=\ln \left (\dfrac{1}{\sqrt{2}} \right )=−\dfrac{1}{2}\ln(2)\) Use properties of exponentsBeing able to solve equations of the form \(y=Ae^{kt}\) suggests a final way of solving exponential equations that can be rewritten in the form \( a = b^{p(x)} \). We will redo example 5 using this alternate method. The method used in example 5 is good practice using log properties. This alternative approach uses exponent properties instead.
Example \(\PageIndex{8}\): Simplify using Exponent Rules before writing in Logarithmic form Solve \(5^{x+2}=4^x\). Solution \[\begin{align*} Exponential Equations that are quadratic in formExtraneous SolutionsSometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the value of the argument of a logarithm is negative, there is no output. Example \(\PageIndex{9}\): Solve Exponential Functions that are Quadratic in Form Solve \(e^{2x}−e^x=56\). Solution: \[\begin{align*} e^{2x}-e^x&= 56\\ Analysis When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation \(e^x=−7\) because a positive number never equals a negative number. The solution \(\ln(−7)\) is not a real number, and in the real number system this solution is rejected as an extraneous solution.
Solve \(e^{2x}=e^x+2\). Answer\(x=\ln2\) Solving Logarithmic Equations
No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions. Rewrite in Exponential FormWe 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 rewrite the logarithmic equation in exponential form to solve for \(x\): \[\begin{align*} {\log}_2(2)+{\log}_2(3x-5)&= 3\\ {\log}_2(2(3x-5))&= 3 \qquad &&\text{Apply the product rule of logarithms}\\ {\log}_2(6x-10)&= 3 \qquad &&\text{Distribute}\\ 2^3&= 6x-10 \qquad &&\text{Apply the definition of a logarithm}\\ 8&= 6x-10 \qquad &&\text{Calculate } 2^3\\ 18&= 6x \qquad &&\text{Add 10 to both sides}\\ x&= 3 \qquad &&\text{Divide by 6, then check the solution!} \end{align*}\] Whenever solving an equation with logarithms, it is always necessary to check that the solution is in the domain of the original equation. If it is not, it must be rejected as a solution. USE THE DEFINITION OF A LOGARITHM TO SOLVE LOGARITHMIC EQUATIONS For any algebraic expression \(S\) and real numbers \(b\) and \(c\), where \(b>0\), \(b≠1\), \[\begin{align} {\log}_b(S)=c \text{ if and only if } b^c=S \end{align}\] Example \(\PageIndex{10}\): Rewrite a Logarithmic Equation in Exponential Form Solve \(2\ln x+3=7\). Solution \[\begin{align*} 2\ln x+3&= 7\\ 2\ln x&= 4 \qquad &&\text{Subtract 3}\\ \ln x&= 2 \qquad &&\text{Divide by 2}\\ x&= e^2 \qquad &&\text{Rewrite in exponential form} \end{align*}\]
Solve \(6+\ln x=10\). Answer\(x=e^4\) Example \(\PageIndex{11}\) Solve \(2\ln(6x)=7\). Solution: \[\begin{align*} 2\ln(6x)&= 7\\ \ln(6x)&= \dfrac{7}{2} \qquad &&\text{Divide by 2}\\ 6x&= e^{\left (\dfrac{7}{2} \right )} \qquad &&\text{Use the definition of }\ln \\ x&= \tfrac{1}{6}e^{\left (\tfrac{7}{2} \right )} &&\qquad \text{Divide by 6} \end{align*}\]
Solve \(2\ln(x+1)=10\). Answer\(x=e^5−1\) Use the One-to-One Property of LogarithmsAs 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 \(S>0\), \(T>0\) 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 \(\log(3x−2)−\log(2)=\log(x+4)\). \[\begin{align*} \log(3(10)-2)-\log(2)&= \log((10)+4) \\ \log(28)-\log(2)&= \log(14)\\ \log \left (\dfrac{28}{2} \right )&= \log(14) \qquad \text{The solution checks} \end{align*}\] USE THE ONE-TO-ONE PROPERTY OF LOGARITHMS TO SOLVE LOGARITHMIC EQUATIONS For any algebraic expressions \(S\) and \(T\) and any positive real number \(b\), where \(b≠1\), \[\begin{align*} {\log}_bS={\log}_bT \quad \text{ if and only if } \quad S=T \end{align*}\] Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.
Example \(\PageIndex{12}\): Use the One-to-One Property of Logarithms to Solve an Equation 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}\\ 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 \(x=−1\) Key Equations
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