How to solve exponential equations

To solve an exponential equation, isolate the exponential term and take the logarithm of both sides. The logarithm undoes the exponent, turning an equation like b^x = c into x = log_b(c). The most common mistake is treating log(x) as a variable or confusing log and ln, they have different bases (10 and e, respectively) and are not interchangeable. Five worked examples cover the main cases you will encounter.

When both sides share the same base, skip the log and match the exponents. For instance, take 4^(x) = 2^(x+3). Rewrite 4 as 2^2, giving (2^2)^x = 2^(x+3), which simplifies to 2^(2x) = 2^(x+3). Since the bases are equal, the exponents must be equal: 2x = x+3, so x = 3. This works whenever you can express both sides as powers of the same base. If you cannot, take logarithms.

When the bases differ, take the logarithm of both sides. The power rule, log_b(M^p) = p * log_b(M), lets you bring the exponent down as a coefficient. For example, solve 3^(2x) = 7. Take log_10 of both sides: log(3^(2x)) = log(7) → 2x * log(3) = log(7) → x = (1/2) * log(7)/log(3). Using a calculator, log(7) ≈ 0.8451 and log(3) ≈ 0.4771, so x ≈ 0.5 * 1.7712 = 0.8856. This is the standard approach for solving exponential equations with logarithms. If your calculator has a logBASE( function, accessible on a TI-84 Plus CE via the MATH key, compute log_3(7) directly.

Example: 2^x = 5

Take log of both sides: log(2^x) = log(5) → x * log(2) = log(5) → x = log(5)/log(2) ≈ 0.6990/0.3010 ≈ 2.3219. Alternatively, use the change of base formula: x = log_2(5).

When the exponential involves Euler's number e, use the natural log ln.For e^(0.1x) = 20, take ln of both sides: ln(e^(0.1x)) = ln(20) → 0.1x = ln(20) → x = 10 * ln(20) ≈ 10 * 2.9957 = 29.957. The same logic applies to any base: for base 10, use log_10; for base 2, use log_2. The key is picking the log that matches the base, so the exponent isolates cleanly.

When both sides have exponentials with different bases, take the log of both sides and apply the power rule. For b^x = d^(x+1), take ln: ln(b^x) = ln(d^(x+1)) → x * ln(b) = (x+1) * ln(d) → x * ln(b) = x * ln(d) + ln(d) → x(ln(b) - ln(d)) = ln(d) → x = ln(d) / (ln(b) - ln(d)). This works for any base; pick a consistent log function, ln or log_10, and proceed.

Example: 4^x = 2^(x+3)

Take log_10: log(4^x) = log(2^(x+3)) → x * log(4) = (x+3) * log(2) → x * log(4) = x * log(2) + 3 * log(2) → x(log(4) - log(2)) = 3 * log(2) → x = (3 * log(2)) / (log(4) - log(2)). Since log(4) = log(2^2) = 2*log(2), this simplifies to x = (3 * log(2)) / (2*log(2) - log(2)) = 3.

Some exponential equations look like quadratics. For instance, e^(2x) - 3e^x + 2 = 0. Let u = e^x. Then the equation becomes u^2 - 3u + 2 = 0, which factors as (u - 1)(u - 2) = 0. So u = 1 or u = 2. Substitute back: e^x = 1 → x = ln(1) = 0; e^x = 2 → x = ln(2) ≈ 0.6931. Always check that the argument of the log is positive: e^x is always positive, so no extraneous solutions arise here. This technique works for any base: 2^(2x) - 5*2^x + 6 = 0 becomes u^2 - 5u + 6 = 0 with u = 2^x.

Exponential equations model doubling time and half-life. For doubling time, a quantity doubles after a fixed interval: if a population follows N = N_0 * 2^(t/T), then when N = 2N_0, you have 2 = 2^(t/T) → t/T = 1 → t = T. For half-life, the equation is N = N_0 * (1/2)^(t/T), and when N = N_0/2, you get 1/2 = (1/2)^(t/T) → t/T = 1 → t = T. These are the simplest cases. More realistic problems give you a specific quantity and ask for the time: for example, how long to reach 5 times the initial amount? Then solve 5 = 2^(t/T) by taking log_2 of both sides.

Example: Doubling Time

An investment doubles every 7 years. How long to reach 10 times the original amount? Solve 10 = 2^(t/7) → take log_2: log_2(10) = t/7 → t = 7 * log_2(10) ≈ 7 * 3.3219 = 23.25 years.

Common Mistakes

The most frequent error is taking log(a+b) and treating it as log(a) + log(b), those are not equal. Another is forgetting that the argument of a log must be positive: if a substitution like u = e^x gives u = -1, discard it because e^x cannot be negative.

This method suits Algebra II and precalculus students who need to understand function inverses, chemistry students working with pH (a negative base-10 log of hydrogen ion concentration), and adults using Excel for financial modeling. It also serves computer science students analyzing algorithm complexity with binary logarithms. If you need the history of Napier or Briggs, or the physical construction of a slide rule, go to a history-of-mathematics site. If you need advanced calculus involving natural log integrals or series expansions, consult a calculus textbook.