Logarithms and Exponential Equations
Aligned to F-LE.A.4 — Common Core State Standards for Mathematics.

What this lesson teaches
A logarithm answers 'what exponent is needed?' It's the inverse of an exponential, so logs solve equations where the variable is an exponent.
Worked examples
Example 1
Solve 2^x = 8. Since 2³ = 8, x = 3. In log form: log₂(8) = 3.
Example 2
Solve 5·3^x = 405. First isolate the exponential: divide by 5 to get 3^x = 81. Since 3^4 = 81, x = 4. In log form, x = log₃(81) = 4. Isolating before taking a log is what makes the step legal.
Example 3
Common mistake: distributing a log across a sum. log(a + b) is NOT log a + log b — log(2 + 8) = log 10 = 1, while log 2 + log 8 = 0.301 + 0.903 = 1.204. The real rules go the other way: log(ab) = log a + log b, log(a/b) = log a − log b, and log(a^n) = n·log a. Only a PRODUCT splits into a sum.
Practice questions
- Rewrite 10² = 100 in log form.
- Solve 3^x = 27.
- Why is a logarithm the inverse of an exponential?
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
Logaritmos y Ecuaciones Exponenciales
Un logaritmo responde a la pregunta '¿qué exponente se necesita?' Es la operación inversa de una exponencial, por lo que los logaritmos sirven para resolver ecuaciones donde la variable es el exponente.
Ejemplo 1: Resuelve 2^x = 8. Como 2³ = 8, entonces x = 3. En forma logarítmica: log₂(8) = 3.
Ejemplo 2: Resuelve 5·3^x = 405. Primero aísla la exponencial: divide entre 5 para obtener 3^x = 81. Como 3^4 = 81, x = 4. En forma logarítmica, x = log₃(81) = 4. Aislar antes de tomar el logaritmo es lo que hace legal el paso.
Ejemplo 3: Error común: distribuir un logaritmo sobre una suma. log(a + b) NO es log a + log b — log(2 + 8) = log 10 = 1, mientras que log 2 + log 8 = 0.301 + 0.903 = 1.204. Las reglas verdaderas van al revés: log(ab) = log a + log b, log(a/b) = log a − log b, y log(a^n) = n·log a. Solo un PRODUCTO se convierte en suma.
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