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Study Guides > College Algebra CoRequisite Course

Exponential Equations

Learning Outcomes

  • Solve an exponential equation with a common base.
  • Rewrite an exponential equation so all terms have a common base then solve.
  • Recognize when an exponential equation does not have a solution.
  • Use logarithms to solve exponential equations.

Exponential Equations

The first technique we will introduce for solving exponential equations involves two functions with like bases. The one-to-one property of exponential functions tells us that, for any real numbers b, S, and T, where [latex]b>0,\text{ }b\ne 1[/latex], [latex]{b}^{S}={b}^{T}[/latex] 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 [latex]{3}^{4x - 7}=\frac{{3}^{2x}}{3}[/latex]. 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:

[latex]\begin{array}{l}{3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{3}\hfill & \hfill \\ {3}^{4x - 7}\hfill & =\frac{{3}^{2x}}{{3}^{1}}\hfill & {\text{Rewrite 3 as 3}}^{1}.\hfill \\ {3}^{4x - 7}\hfill & ={3}^{2x - 1}\hfill & \text{Use the division property of exponents}\text{.}\hfill \\ 4x - 7\hfill & =2x - 1\text{ }\hfill & \text{Apply the one-to-one property of exponents}\text{.}\hfill \\ 2x\hfill & =6\hfill & \text{Subtract 2}x\text{ and add 7 to both sides}\text{.}\hfill \\ x\hfill & =3\hfill & \text{Divide by 3}\text{.}\hfill \end{array}[/latex]

tip for success

Note the use of the exponent rules and the one-to-one property of exponents in the example above. Return to the review section for this module or to Algebra Essentials anytime if you need a refresher.

A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations

For any algebraic expressions S and T, and any positive real number [latex]b\ne 1[/latex], [latex-display]{b}^{S}={b}^{T}\text{ if and only if }S=T[/latex-display]

How To: Given an exponential equation Of the form [latex]{b}^{S}={b}^{T}[/latex], where S and T are algebraic expressions with an unknown, solve for the unknown

  1. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[/latex].
  2. Use the one-to-one property to set the exponents equal to each other.
  3. Solve the resulting equation, S = T, for the unknown.

Example: Solving an Exponential Equation with a Common Base

Solve [latex]{2}^{x - 1}={2}^{2x - 4}[/latex].

Answer: [latex-display]\begin{array}{l} {2}^{x - 1}={2}^{2x - 4}\hfill & \text{The common base is }2.\hfill \\ \text{ }x - 1=2x - 4\hfill & \text{By the one-to-one property the exponents must be equal}.\hfill \\ \text{ }x=3\hfill & \text{Solve for }x.\hfill \end{array}[/latex-display]

Try It

Solve [latex]{5}^{2x}={5}^{3x+2}[/latex].

Answer: [latex]x=–2[/latex]

[ohm_question]2637[/ohm_question]

Rewriting Equations So All Powers Have the Same Base

Sometimes 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 [latex]256={4}^{x - 5}[/latex]. 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:

[latex]\begin{array}{l}256={4}^{x - 5}\hfill & \hfill \\ {2}^{8}={\left({2}^{2}\right)}^{x - 5}\hfill & \text{Rewrite each side as a power with base 2}.\hfill \\ {2}^{8}={2}^{2x - 10}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ 8=2x - 10\hfill & \text{Apply the one-to-one property of exponents}.\hfill \\ 18=2x\hfill & \text{Add 10 to both sides}.\hfill \\ x=9\hfill & \text{Divide by 2}.\hfill \end{array}[/latex]

How To: Given an exponential equation with unlike bases, use the one-to-one property to solve it

  1. Rewrite each side in the equation as a power with a common base.
  2. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form [latex]{b}^{S}={b}^{T}[/latex].
  3. Use the one-to-one property to set the exponents equal to each other.
  4. Solve the resulting equation, S = T, for the unknown.

Example: Solving Equations by Rewriting Them to Have a Common Base

Solve [latex]{8}^{x+2}={16}^{x+1}[/latex].

Answer: [latex-display]\begin{array}{llllll}\text{ }{8}^{x+2}={16}^{x+1}\hfill & \hfill \\ {\left({2}^{3}\right)}^{x+2}={\left({2}^{4}\right)}^{x+1}\hfill & \text{Write }8\text{ and }16\text{ as powers of }2.\hfill \\ \text{ }{2}^{3x+6}={2}^{4x+4}\hfill & \text{To take a power of a power, multiply the exponents}.\hfill \\ \text{ }3x+6=4x+4\hfill & \text{Use the one-to-one property to set the exponents equal to each other}.\hfill \\ \text{ }x=2\hfill & \text{Solve for }x.\hfill \end{array}[/latex-display]

Try It

Solve [latex]{5}^{2x}={25}^{3x+2}[/latex].

Answer: [latex]x=–1[/latex]

[ohm_question]2620[/ohm_question]

Example: Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve [latex]{2}^{5x}=\sqrt{2}[/latex].

Answer: [latex-display]\begin{array}{l}{2}^{5x}={2}^{\frac{1}{2}}\hfill & \text{Write the square root of 2 as a power of }2.\hfill \\ 5x=\frac{1}{2}\hfill & \text{Use the one-to-one property}.\hfill \\ x=\frac{1}{10}\hfill & \text{Solve for }x.\hfill \end{array}[/latex-display]

Try It

Solve [latex]{5}^{x}=\sqrt{5}[/latex].

Answer: [latex]x=\frac{1}{2}[/latex]

[ohm_question]2638[/ohm_question]

Q & A

Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process? No. Recall that the range of an exponential function is always positive. While solving the equation we may obtain an expression that is undefined.

Example: Determine When an Equation has No Solution

Solve [latex]{3}^{x+1}=-2[/latex].

Answer: 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.

Analysis of the Solution

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 of the equation. Thus the equation has no solution. Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.

Try It

Solve [latex]{2}^{x}=-100[/latex].

Answer: The equation has no solution.

[ohm_question]98554[/ohm_question]

Using Logarithms to Solve Exponential Equations

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 that since [latex]\mathrm{log}\left(a\right)=\mathrm{log}\left(b\right)[/latex] is equivalent to a = b, we may apply logarithms with the same base to both sides of an exponential equation.

applying the one-to-one property of logarithms

Note in the paragraph above the reiteration of the one-to-one property of logarithms. When applying logarithms with the same base to both sides of an exponential equation, we often use the common logarithm, [latex]\log[/latex] or the natural logarithm, [latex]\ln[/latex]. The choice is yours which to use in most situations, but if either base in a given exponential equation is [latex]10[/latex], use [latex]\log[/latex] or if the base is [latex]e[/latex], use [latex]\ln[/latex] to take advantage of the identity property of logarithms.

How To: Given an exponential equation Where a common base cannot be found, solve for the unknown

  1. Apply the logarithm to 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.
  2. Use the rules of logarithms to solve for the unknown.

Example: Solving an Equation Containing Powers of Different Bases

Solve [latex]{5}^{x+2}={4}^{x}[/latex].

Answer: [latex-display]\begin{array}{l}\text{ }{5}^{x+2}={4}^{x}\hfill & \text{There is no easy way to get the powers to have the same base}.\hfill \\ \text{ }\mathrm{ln}{5}^{x+2}=\mathrm{ln}{4}^{x}\hfill & \text{Take ln of both sides}.\hfill \\ \text{ }\left(x+2\right)\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the power rule for logs}.\hfill \\ \text{ }x\mathrm{ln}5+2\mathrm{ln}5=x\mathrm{ln}4\hfill & \text{Use the distributive property}.\hfill \\ \text{ }x\mathrm{ln}5-x\mathrm{ln}4=-2\mathrm{ln}5\hfill & \text{Get terms containing }x\text{ on one side, terms without }x\text{ on the other}.\hfill \\ x\left(\mathrm{ln}5-\mathrm{ln}4\right)=-2\mathrm{ln}5\hfill & \text{On the left hand side, factor out }x.\hfill \\ \text{ }x\mathrm{ln}\left(\frac{5}{4}\right)=\mathrm{ln}\left(\frac{1}{25}\right)\hfill & \text{Use the properties of logs}.\hfill \\ \text{ }x=\frac{\mathrm{ln}\left(\frac{1}{25}\right)}{\mathrm{ln}\left(\frac{5}{4}\right)}\hfill & \text{Divide by the coefficient of }x.\hfill \end{array}[/latex-display]

Try It

Solve [latex]{2}^{x}={3}^{x+1}[/latex].

Answer: [latex]x=\frac{\mathrm{ln}3}{\mathrm{ln}\left(\frac{2}{3}\right)}[/latex]

[ohm_question]98555[/ohm_question]

Q & A

Is there any way to solve [latex]{2}^{x}={3}^{x}[/latex]? Yes. The solution is x = 0.

Equations Containing [latex]e[/latex]

One common type of exponential equations are those with base e. This constant occurs again and again in nature, mathematics, science, engineering, and finance. When we have an equation with a base e on either side, we can use the natural logarithm to solve it.

How To: Given an equation of the form [latex]y=A{e}^{kt}[/latex], solve for [latex]t[/latex]

  1. Divide both sides of the equation by A.
  2. Apply the natural logarithm to both sides of the equation.
  3. Divide both sides of the equation by k.

Example: Solve an Equation of the Form [latex]y=A{e}^{kt}[/latex]

Solve [latex]100=20{e}^{2t}[/latex].

Answer: [latex-display]\begin{array}{l}100\hfill & =20{e}^{2t}\hfill & \hfill \\ 5\hfill & ={e}^{2t}\hfill & \text{Divide by the coefficient 20}\text{.}\hfill \\ \mathrm{ln}5\hfill & =\mathrm{ln}{{e}^{2t}}\hfill & \text{Take ln of both sides.}\hfill \\ \mathrm{ln}5\hfill & =2t\hfill & \text{Use the fact that }\mathrm{ln}\left(x\right)\text{ and }{e}^{x}\text{ are inverse functions}\text{.}\hfill \\ t\hfill & =\frac{\mathrm{ln}5}{2}\hfill & \text{Divide by the coefficient of }t\text{.}\hfill \end{array}[/latex-display]  

Analysis of the Solution

Using laws of logs, we can also write this answer in the form [latex]t=\mathrm{ln}\sqrt{5}[/latex]. If we want a decimal approximation of the answer, then we use a calculator.

tip for success

Just as you have done when solving various types of equations, isolate the term containing the variable for which you are solving before applying any properties of equality or inverse operations. That's why, in the example above, you must divide away the [latex]A[/latex] first.

Try It

Solve [latex]3{e}^{0.5t}=11[/latex].

Answer: [latex]t=2\mathrm{ln}\left(\frac{11}{3}\right)[/latex] or [latex]\mathrm{ln}{\left(\frac{11}{3}\right)}^{2}[/latex]

[ohm_question]98596[/ohm_question]

Q & A

Does every equation of the form [latex]y=A{e}^{kt}[/latex] have a solution? No. There is a solution when [latex]k\ne 0[/latex], and when [latex]y[/latex] and [latex]</em>A<em>[/latex] are either both 0 or neither 0 and they have the same sign. An example of an equation with this form that has no solution is [latex]2=-3{e}^{t}[/latex].

Example: Solving an Equation That Can Be Simplified to the Form [latex]y=A{e}^{kt}[/latex]

Solve [latex]4{e}^{2x}+5=12[/latex].

Answer: [latex-display]\begin{array}{l}4{e}^{2x}+5=12\hfill & \hfill \\ 4{e}^{2x}=7\hfill & \text{Subtract 5 from both sides}.\hfill \\ {e}^{2x}=\frac{7}{4}\hfill & \text{Divide both sides by 4}.\hfill \\ 2x=\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Take ln of both sides}.\hfill \\ x=\frac{1}{2}\mathrm{ln}\left(\frac{7}{4}\right)\hfill & \text{Solve for }x.\hfill \end{array}[/latex-display]

Try It

Solve [latex]3+{e}^{2t}=7{e}^{2t}[/latex].

Answer: [latex]t=\mathrm{ln}\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{2}\mathrm{ln}\left(2\right)[/latex]

[ohm_question]129891[/ohm_question]

Extraneous Solutions

Sometimes 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 taking the logarithm of both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

tip for success

Always check solutions to logarithmic equations to see if they satisfy the domain restriction on the logarithm.

Example: Solving Exponential Functions in Quadratic Form

Solve [latex]{e}^{2x}-{e}^{x}=56[/latex].

Answer:   [latex-display]\begin{array}{l}{e}^{2x}-{e}^{x}=56\hfill \\ {e}^{2x}-{e}^{x}-56=0\hfill & \text{Get one side of the equation equal to zero}.\hfill \\ \left({e}^{x}+7\right)\left({e}^{x}-8\right)=0\hfill & \text{Factor by the FOIL method}.\hfill \\ {e}^{x}+7=0\text{ or }{e}^{x}-8=0 & \text{If a product is zero, then one factor must be zero}.\hfill \\ {e}^{x}=-7{\text{ or e}}^{x}=8\hfill & \text{Isolate the exponentials}.\hfill \\ {e}^{x}=8\hfill & \text{Reject the equation in which the power equals a negative number}.\hfill \\ x=\mathrm{ln}8\hfill & \text{Solve the equation in which the power equals a positive number}.\hfill \end{array}[/latex-display]

Analysis of the Solution

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 [latex]{e}^{x}=-7[/latex] because a positive number never equals a negative number. The solution [latex]x=\mathrm{ln}\left(-7\right)[/latex] is not a real number and in the real number system, this solution is rejected as an extraneous solution.

Try It

Solve [latex]{e}^{2x}={e}^{x}+2[/latex].

Answer: [latex]x=\mathrm{ln}2[/latex]

[ohm_question]98598[/ohm_question]

Q & A

Does every logarithmic equation have a solution? No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

Licenses & Attributions

CC licensed content, Original

CC licensed content, Shared previously

  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay et al.. Located at: https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites. License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
  • Question ID 2637, 2620, 2638. Authored by: Langkamp,Greg. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.
  • Question ID 98554, 98555, 98596. Authored by: Jenck,Michael. License: CC BY: Attribution. License terms: IMathAS Community License CC-BY + GPL.