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Change of Base & Solving Exponential Equations

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Every logarithm is a ratio of two others

A calculator provides log (base 10) and ln (base e), but log₇(200) is a perfectly reasonable thing to want. The change-of-base formula builds it from either one: log_b(x) = log(x)/log(b) = ln(x)/ln(b). The choice of base for the two logs on the right is free, as long as it is the same in numerator and denominator — the ratio comes out the same either way.

Change of base
log_b(x) = ln(x)/ln(b) = log(x)/log(b) — any common base works
This is a ratio, not a difference. log(x)/log(b) is not log(x) − log(b), and confusing the two is the standard error.
Worked example

Evaluate log₇(200) to four decimal places.

  1. 1.Apply change of base: log₇(200) = ln(200)/ln(7).
  2. 2.ln(200) ≈ 5.2983 and ln(7) ≈ 1.9459.
  3. 3.5.2983/1.9459 ≈ 2.7229.
  4. 4.Sanity check: 7² = 49 and 7³ = 343, and 200 lies between them, so the answer must lie between 2 and 3. ✓
Answer: log₇(200) ≈ 2.7229. The bracketing check — 49 < 200 < 343, so the answer is between 2 and 3 — catches almost every calculator slip, including the difference-instead-of-ratio error, which would give 5.2983 − 1.9459 ≈ 3.35.

Two routes for an exponential equation

When the unknown is in an exponent, there are two strategies. If both sides can be written with the same base, equate the exponents — exact, fast, no logarithms. If they cannot, take a logarithm of both sides and use the power rule log(a^x) = x·log(a) to pull the unknown down out of the exponent. Try same-base first; it is cleaner and yields exact answers.

Worked example

Solve 2^(3x−1) = 32.

  1. 1.Recognize 32 = 2⁵, so both sides share base 2.
  2. 2.2^(3x−1) = 2⁵.
  3. 3.An exponential function is one-to-one, so equal outputs force equal exponents: 3x − 1 = 5.
  4. 4.3x = 6, so x = 2.
  5. 5.Check: 2^(3(2)−1) = 2⁵ = 32. ✓
Answer: x = 2, obtained exactly and without logarithms. Matching bases works whenever both sides are powers of a common number — always scan for it before reaching for ln.
Worked example

Solve 5·3^(2x) = 400, giving an exact expression and a decimal to three places.

  1. 1.Isolate the exponential first: 3^(2x) = 80.
  2. 2.Take the natural log of both sides: ln(3^(2x)) = ln(80).
  3. 3.Power rule brings the exponent down: 2x·ln(3) = ln(80).
  4. 4.Solve: x = ln(80)/(2·ln(3)).
  5. 5.Numerically: ln(80) ≈ 4.3820 and ln(3) ≈ 1.0986, so x ≈ 4.3820/2.1972 ≈ 1.994.
Answer: x = ln(80)/(2 ln 3) exactly, or x ≈ 1.994. Isolating the exponential before taking logs is essential — logging 5·3^(2x) directly gives ln 5 + 2x ln 3, which is still solvable but invites algebra errors.
Watch out

ln(5·3^(2x)) is not 5·ln(3^(2x)). A coefficient multiplying an exponential becomes an added logarithm, not a multiplied one: ln(5) + 2x·ln(3). Dividing the 5 out first avoids the issue entirely.

Checkpoint

Solve 4^(x+1) = 8^(x−2).

When the exponential appears twice

An equation like 4^x − 5·2^x + 4 = 0 looks unfamiliar until you notice 4^x = (2²)^x = (2^x)². Substituting u = 2^x turns it into the quadratic u² − 5u + 4 = 0, with roots u = 1 and u = 4. Then back-substitute: 2^x = 1 gives x = 0, and 2^x = 4 gives x = 2. This substitution trick handles any equation that is quadratic in a single exponential.

Checkpoint

In solving 9^x − 4·3^x + 3 = 0 by substituting u = 3^x, you find u = 1 and u = 3. What are the solutions for x?

Answer the 2 checkpoints as you read.

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