6.14: Solving Exponential Equations

Definitions and Theorems

An exponential equation is solved by undoing the exponent. Two tools make this possible: rewriting both sides so they share a common base, or applying a logarithm—the inverse of exponentiation—to both sides. This section reviews both strategies.

Definition: Exponential Equation

An exponential equation is an equation in which a variable appears in an exponent.

Definition: Logarithm

Let \(b>0\) with \(b\neq 1\), and let \(x>0\). The logarithm of \(x\) to base \(b\), written \(\log_b(x)\), is the exponent to which \(b\) must be raised to produce \(x\). That is, for any real number \(y\),\[y=\log_b(x)\quad\Longleftrightarrow\quad b^{y}=x.\nonumber\]

This equivalence is the inverse relationship between exponentiation and logarithms: the logarithm base \(b\) reverses raising \(b\) to a power. It is the relationship that lets you extract a variable trapped in an exponent.

Notation: Common and Natural Logarithms

Two bases occur often enough to have their own symbols. The common logarithm is base \(10\), written \(\log(x)=\log_{10}(x)\). The natural logarithm is base \(e\) (where \(e\approx 2.718\)), written \(\ln(x)=\log_e(x)\). Applying the inverse relationship to these gives the identities\[b^{\log_b(x)}=x\qquad\text{and}\qquad\log_b\!\left(b^{x}\right)=x,\nonumber\]and, in particular, \(\ln\!\left(e^{x}\right)=x\).

Theorem: One-to-One Property of Exponents

For any algebraic expressions \(S\) and \(T\), and any positive real number \(b \neq 1\),\[b^S = b^T \text{ if and only if } S = T. \nonumber \]

Proof

Let \( S \) and \(T\) be algebraic expressions, and \( b \) be a positive real number not equal to 1.\[ \begin{array}{rrclcl}
& b^S & = & b^T & & \\[6pt]
\iff & \ln(b^S) & = & \ln(b^T) & \quad & \left( \text{taking the natural log of both sides} \right) \\[6pt]
\iff & S \ln(b) & = & T \ln(b) & \quad & \left( \text{Laws of Logarithms} \right) \\[6pt]
\iff & S & = & T & \quad & \left( \text{dividing both sides by }\ln(b) \right) \\[6pt]
\end{array}  \nonumber \]

This property is the basis for the equating-bases method: if both sides of an equation can be written as powers of the same base, the exponents must be equal.

Theorem: One-to-One Property of Logarithms

Let \(b>0\) with \(b\neq 1\), and let \(x>0\) and \(y>0\). Then\[x=y\quad\Longleftrightarrow\quad\log_b(x)=\log_b(y).\nonumber\]

This property justifies taking the logarithm of both sides: applying the same logarithm to two equal positive quantities preserves the equality, and no solutions are gained or lost.

Theorem: Power Rule for Logarithms

Let \(b>0\) with \(b\neq 1\), let \(x>0\), and let \(p\) be any real number. Then\[\log_b\!\left(x^{p}\right)=p\log_b(x).\nonumber\]

The power rule is what makes logarithms useful here: it moves an exponent down into a coefficient, turning an unknown exponent into an unknown factor that ordinary algebra can isolate.

Choosing a Logarithm

By the one-to-one property, you may take a logarithm to any base of both sides. Base \(10\) (\(\log\)) and base \(e\) (\(\ln\)) are the usual choices because calculators evaluate them directly. When the base of the equation is \(e\), prefer \(\ln\), since \(\ln\!\left(e^{x}\right)=x\) collapses the left side immediately.

Caution: Where the Exponent Lands

The power rule brings the exponent out as a factor of the whole logarithm: \(\log_b\!\left(x^{p}\right)=p\log_b(x)\). It does not raise the logarithm to a power; that is, \(\log_b\!\left(x^{p}\right)\neq\left(\log_b(x)\right)^{p}\). Also, equating bases is only valid once both sides are written as powers of the same base—if no common base is available, take a logarithm of both sides instead.

Examples

Example \(\PageIndex{1}\): Equating Bases

Solve \(2^{x}=32\).

Answer

Write the right side as a power of \(2\), since \(32=2^{5}\). With a common base, the one-to-one property lets you equate exponents.\[\begin{aligned}2^{x}&=32\\2^{x}&=2^{5}\\x&=5.\end{aligned}\nonumber\]

Example \(\PageIndex{2}\): A Common Base After Rewriting

Solve \(8^{x+1}=32\).

Answer

Both \(8\) and \(32\) are powers of \(2\): \(8=2^{3}\) and \(32=2^{5}\). Rewrite each side, combine the exponents with the power-of-a-power rule, then equate exponents.\[\begin{aligned}8^{x+1}&=32\\\left(2^{3}\right)^{x+1}&=2^{5}\\2^{3(x+1)}&=2^{5}\\3(x+1)&=5\\3x+3&=5\\x&=\dfrac{2}{3}.\end{aligned}\nonumber\]

Example \(\PageIndex{3}\): No Common Base

Solve \(5^{x}=40\). Give an exact answer and a decimal approximation to four places.

Answer

Since \(40\) is not a power of \(5\), take the common logarithm of both sides, then use the power rule \(\log\!\left(5^{x}\right)=x\log 5\) to free the exponent.\[\begin{aligned}5^{x}&=40\\\log\!\left(5^{x}\right)&=\log 40\\x\log 5&=\log 40\\x&=\dfrac{\log 40}{\log 5}\approx 2.2920.\end{aligned}\nonumber\]

Example \(\PageIndex{4}\): Base \(e\)

Solve \(e^{2x}=15\). Give an exact answer and a decimal approximation to four places.

Answer

The base is \(e\), so take the natural logarithm of both sides; the left side collapses because \(\ln\!\left(e^{2x}\right)=2x\).\[\begin{aligned}e^{2x}&=15\\\ln\!\left(e^{2x}\right)&=\ln 15\\2x&=\ln 15\\x&=\dfrac{\ln 15}{2}\approx 1.3540.\end{aligned}\nonumber\]

Example \(\PageIndex{5}\): An Exponent Expression

Solve \(2^{x+1}=7\). Give an exact answer and a decimal approximation to four places.

Answer

Take the common logarithm of both sides. The power rule brings the entire exponent \(x+1\) down as a factor; then isolate \(x\).\[\begin{aligned}2^{x+1}&=7\\(x+1)\log 2&=\log 7\\x+1&=\dfrac{\log 7}{\log 2}\\x&=\dfrac{\log 7}{\log 2}-1\approx 1.8074.\end{aligned}\nonumber\]

Example \(\PageIndex{6}\): Different Bases on Each Side

Solve \(3^{2x-1}=5^{x+2}\). Give an exact answer and a decimal approximation to four places.

Answer

No single base serves both sides, so take the natural logarithm of both sides and apply the power rule to each exponent. Distribute, gather the \(x\) terms on one side, and factor.\[\begin{aligned}3^{2x-1}&=5^{x+2}\\(2x-1)\ln 3&=(x+2)\ln 5\\2x\ln 3-\ln 3&=x\ln 5+2\ln 5\\2x\ln 3-x\ln 5&=2\ln 5+\ln 3\\x(2\ln 3-\ln 5)&=2\ln 5+\ln 3\\x&=\dfrac{2\ln 5+\ln 3}{2\ln 3-\ln 5}\approx 7.3453.\end{aligned}\nonumber\]

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Sources

Several parts of this text use modifications from the following sources:

• Wikipedia article: "Logarithm"
• Wikipedia article: "Natural logarithm"

All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.