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\( \log_b x \)

Solving Exponential
Equations with Logarithms

Learn to unlock unknown exponents by applying logarithms to both sides of an equation: transform \( ab^{ct} = d \) into a clean algebraic solution.

Algebra 2

What Is an Exponential Equation?

An exponential equation is any equation where the unknown variable appears in the exponent. For example:

Because the variable is "trapped" in the exponent, we cannot solve by simple algebra. We need a special tool: logarithms.

The Logarithm: Inverse of Exponentiation

A logarithm answers the question: "To what power must the base be raised to produce this value?"

\( \log_{10} \)
Common log (base 10). Written as \( \log x \).
\( \ln \)
Natural log (base \( e \approx 2.718 \)).
\( \log_{2} \)
Binary log (base 2). Common in CS and doubling problems.

Visualizing the Exponential-Log Connection

The Core Rule: Taking the Log of Both Sides

For any equation of the form \( ab^{ct} = d \):

1
Isolate the exponential term: divide both sides by \( a \) to get \( b^{ct} = \dfrac{d}{a} \).
2
Apply the logarithm (base \( b \)) to both sides: \( ct = \log_{b}\!\left(\dfrac{d}{a}\right) \).
3
Solve for \( t \): divide by \( c \) to get \( t = \dfrac{1}{c}\,\log_{b}\!\left(\dfrac{d}{a}\right) \).

Why Does This Work?

Logarithms are the inverse of exponentiation. Applying \( \log_b \) to \( b^{(\text{something})} \) "cancels" the base and releases the exponent:

This is exactly like how taking a square root undoes squaring, or how dividing undoes multiplication. The log "undoes" the exponential.

Choosing Your Log Base

You can use any base of logarithm to solve an exponential equation. However, some choices simplify the work:

Match the Base
If the base is 2, 10, or \( e \), use \( \log_2 \), \( \log_{10} \), or \( \ln \) to cancel directly.
Use \( \ln \) or \( \log \)
For bases not on your calculator, use the Power Rule with \( \ln \) or \( \log_{10} \) and divide.
Change of Base
\( \log_b(x) = \dfrac{\ln x}{\ln b} = \dfrac{\log x}{\log b} \). Converts any log to your calculator's keys.

\( \log_b \) vs \( \ln \) vs \( \log \)

Notation Base When to Use Calculator Key
\( \log x \) 10 Base-10 exponentials, pH, decibels LOG
\( \ln x \) \( e \approx 2.718 \) Continuous growth/decay, \( e^{kt} \) LN
\( \log_2 x \) 2 Doubling/halving, binary, half-life \( \frac{\ln x}{\ln 2} \)

Step-by-Step Method

1
Isolate the exponential: Move all non-exponential factors to the other side (divide by the coefficient \( a \)).
2
Take the logarithm: Apply \( \log_b \), \( \ln \), or \( \log \) to both sides of the equation.
3
Bring down the exponent: Use the rule \( \log(b^{y}) = y \cdot \log(b) \) or direct cancellation if the log base matches.
4
Solve for the variable: Use basic algebra (divide, subtract) to isolate \( t \) or \( x \).
5
Evaluate with technology: Use a calculator to find a decimal approximation if needed.

Example 1: Base 10

1
Isolate: Divide both sides by 5: \( 10^{2x} = 600 \)
2
Take \( \log_{10} \): \( \log(10^{2x}) = \log(600) \)
3
Simplify: \( 2x = \log(600) \) because \( \log(10^{2x}) = 2x \).
4
Solve: \( x = \dfrac{\log(600)}{2} \approx \dfrac{2.778}{2} \approx \) 1.389

Example 2: Base \( e \)

1
Isolate: Divide both sides by 200: \( e^{0.03t} = 4 \)
2
Take \( \ln \): \( \ln(e^{0.03t}) = \ln(4) \)
3
Simplify: \( 0.03t = \ln(4) \) because \( \ln(e^{0.03t}) = 0.03t \).
4
Solve: \( t = \dfrac{\ln(4)}{0.03} \approx \dfrac{1.386}{0.03} \approx \) 46.21

Example 3: Base 2

1
Isolate: Divide both sides by 3: \( 2^{5t} = 32 \)
2
Recognize: \( 32 = 2^5 \), so \( 2^{5t} = 2^5 \).
3
Equate exponents: \( 5t = 5 \), giving \( t = \) 1.

Example 4: When the Answer Isn't Clean

1
Isolate: \( 2^{0.5t} = \dfrac{50}{7} \approx 7.143 \)
2
Take \( \log_2 \): \( 0.5t = \log_2(7.143) \)
3
Change of base: \( \log_2(7.143) = \dfrac{\ln(7.143)}{\ln(2)} \approx \dfrac{1.966}{0.693} \approx 2.837 \)
4
Solve: \( t = \dfrac{2.837}{0.5} \approx \) 5.674

The General Formula

For any equation of the form \( ab^{ct} = d \), where \( a, c, d \) are known numbers and \( b \) is the base:

Or equivalently, using any logarithm (via the power rule):

Both forms are equivalent. Use whichever your calculator supports most easily.

Common Mistakes to Avoid

Forgetting to Isolate First
Taking the log before dividing by \( a \). You must isolate \( b^{ct} \) before applying the logarithm.
Log of a Sum
\( \log(A + B) \neq \log A + \log B \). The log of a sum has no simple rule. Only \( \log(AB) = \log A + \log B \).
Wrong Base on Calculator
Using \( \ln \) when you needed \( \log_{10} \), or vice versa, and forgetting the change of base conversion.

Summary Table

Equation Form Log to Apply Solution
\( a \cdot 10^{ct} = d \) \( \log_{10} \) \( t = \dfrac{\log(d/a)}{c} \)
\( a \cdot e^{ct} = d \) \( \ln \) \( t = \dfrac{\ln(d/a)}{c} \)
\( a \cdot 2^{ct} = d \) \( \log_2 \) \( t = \dfrac{\log_2(d/a)}{c} \)
\( a \cdot b^{ct} = d \) Any: \( \ln \) or \( \log \) \( t = \dfrac{\ln(d/a)}{c\ln b} \)

Question 1 of 5

Solve for \( x \): \( 10^{x} = 500 \)

Question 2 of 5

Solve for \( t \): \( 4e^{0.1t} = 20 \)

Question 3 of 5

Which expression equals the solution to \( 6 \cdot 2^{3t} = 48 \)?

Question 4 of 5

An investment follows \( A = 1000\,e^{0.05t} \). How long until \( A = 3000 \)?

Question 5 of 5

Using the change of base formula, \( \log_2(10) \) is equal to: