Exponents and Logarithms

How to solve exponential equations with logarithms?

Quick answer

You can solve an exponential equation by isolating the exponential term, taking the common or natural logarithm of both sides, and applying the power property to bring the exponent down as a multiplier. Next, divide by the logarithm to isolate the variable, and evaluate the final value on a calculator, rounding to four decimal places.

The topic explained

Why logarithms unlock exponential equations

When an unknown variable is trapped in an exponent, ordinary arithmetic cannot pull it down. Applying logarithms to these equations provides a reliable method for this exact problem. Because logarithms are the inverse of exponents, taking the log of both sides allows you to apply the power rule: log(b^x) = x · log(b). This brings the variable down to the main line where you can isolate it with simple division.

You can use either the common logarithm (base 10, written log) or the natural logarithm (base e, written ln). Both yield identical numerical solutions when evaluated on a calculator.

Step-by-step method for solving

First, isolate the exponential expression on one side of the equation. For example, if you have 4 · 3^(2x) = 36, divide both sides by 4 to get 3^(2x) = 9. Second, take the logarithm of both sides: ln(3^(2x)) = ln(9). Third, move the exponent in front: 2x · ln(3) = ln(9). Finally, divide both sides by 2 · ln(3) to find x = ln(9) / (2 · ln(3)) = 1.

If the equation has different exponential bases on each side, such as 2^(x+1) = 5^x, take the logarithm of both sides immediately. Use the distributive property: (x + 1) ln(2) = x ln(5), expand to x ln(2) + ln(2) = x ln(5), collect all x terms on one side, factor out x, and divide.

A real-world growth example

Consider an investment of $1,000 growing at an annual interest rate of 6% compounded continuously, modeled by A = 1000 · e^(0.06t). To find how long it takes for the account to reach $2,500, set up the equation 2500 = 1000 · e^(0.06t).

Divide both sides by 1000 to get 2.5 = e^(0.06t). Because the base is e, take the natural logarithm of both sides: ln(2.5) = ln(e^(0.06t)). Since ln(e^u) = u, the equation simplifies to ln(2.5) = 0.06t. Dividing ln(2.5) by 0.06 gives t ≈ 15.27 years.

Common pitfalls and domain checks

Always check your solutions when dealing with logarithmic equations derived from exponential models. The argument of a logarithm must strictly be greater than zero. If solving produces a value that makes any logarithmic input zero or negative, that solution is extraneous and must be discarded.

Free printable worksheet

Solving Exponential Equations with Logarithms Practice

Intended for 11th grade math students. Solve each equation using logarithms, giving exact expressions or rounded decimal values.

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Solving Exponential Equations with Logarithms PracticeName:

Solve 2^x = 10 for x, rounded to three decimal places.
Solve 5^(x - 1) = 125 using logarithms.
Solve 3 · 4^x = 48.
Solve e^(0.05t) = 3 for t, rounded to three decimal places.
Solve 7^(2x) = 50 for x, rounded to three decimal places.
Solve 3^(x + 2) = 8^x for x, rounded to three decimal places.
A culture of 200 bacteria doubles every hour according to P(t) = 200 · 2^t. In how many hours will the population reach 5,000? Round to three decimal places.

Common questions

Should I use common log or natural log to solve exponential equations?

You can use either common log (base 10) or natural log (base e). Both yield the exact same numerical answer due to the change of base formula. If your equation contains the base e, using the natural log saves time because ln(e) equals 1.

What is an extraneous solution in logarithmic equations?

An extraneous solution is a numerical result produced by correct algebra that does not work in the original expression. Because logarithms are defined only for positive numbers, any value that creates a non-positive logarithmic argument must be rejected.

How do I solve an equation when both sides have different bases and exponents?

Take the logarithm of both sides, bring exponents down as multipliers, and distribute the log terms. Next, collect all terms with the variable on one side, factor out the common variable, and divide by the remaining coefficient expression.

Why can't I take the log of each term before isolating the exponential part?

The logarithm of a sum or product is not the sum or product of individual logarithms. For example, log(a + b) does not equal log(a) + log(b). You must isolate the single exponential term before taking the logarithm of both sides.

What math skills should I review before starting this topic?

Make sure you are comfortable with basic exponent rules, solving linear equations, and the fundamental log rules: the product rule, the quotient rule, and the power rule.

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