Solving Exponential Equations with Logarithms
Aligned to F-LE.A.4 — Common Core State Standards for Mathematics.
What this lesson teaches
To solve a × b^(ct) = d for t, first isolate the power, then take a logarithm: t = log_b(d/a) / c. A calculator evaluates the logarithm.
Worked examples
Example 1
Solve 5 × 2^t = 40. Divide both sides by 5 to get 2^t = 8. Then t = log_2(8) = 3, since 2^3 = 8.
Example 2
A $3,000 investment grows 6% per year. When does it reach $5,000? Solve 3000(1.06)^t = 5000, so (1.06)^t = 5/3. Take the log of both sides: t·log(1.06) = log(1.6667), so t = 0.2218/0.0253 which is about 8.8 years.
Example 3
Common mistake: dividing by the base instead of taking a log. From 2^x = 8, writing x = 8/2 = 4 happens to look plausible but is wrong — the answer is 3, because the exponent is what you are solving for, and division cannot reach it. The only way down from an exponent is a logarithm (or recognizing the power outright).
Practice questions
- Solve 3^t = 81 by writing t as a logarithm.
- Solve 100 × 10^t = 100000 for t.
- Express the solution of 4 × 3^(2t) = 36 as a logarithm, then evaluate it.
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
Resolver ecuaciones exponenciales con logaritmos
Para resolver a × b^(ct) = d en t, primero aísla la potencia, luego aplica un logaritmo: t = log_b(d/a) / c. Una calculadora evalúa el logaritmo.
Ejemplo 1: Resuelve 5 × 2^t = 40. Divide ambos lados entre 5 para obtener 2^t = 8. Entonces t = log_2(8) = 3, ya que 2^3 = 8.
Ejemplo 2: Una inversión de $3,000 crece 6% por año. ¿Cuándo llega a $5,000? Resuelve 3000(1.06)^t = 5000, así que (1.06)^t = 5/3. Toma el logaritmo de ambos lados: t·log(1.06) = log(1.6667), entonces t = 0.2218/0.0253, unos 8.8 años.
Ejemplo 3: Error común: dividir entre la base en vez de tomar un logaritmo. De 2^x = 8, escribir x = 8/2 = 4 parece plausible pero está mal — la respuesta es 3, porque el exponente es lo que se busca, y la división no puede alcanzarlo. La única manera de bajar de un exponente es un logaritmo (o reconocer la potencia de inmediato).
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