Guides / Logarithms

Solving exponential equations: same base, or take a log

When x is up in an exponent, the whole job is getting it back down. There are two ways to do that, and choosing between them takes one look at the bases.

Why it works

Two facts do all the work.

• Same base, same exponent. For a base b > 0 with b ≠ 1, every exponent gives a different output: 2³ = 8, and no other power of 2 equals 8. So if b^u = b^v, then u = v.
• Logs turn exponents into multipliers. If two positive numbers are equal, their logs are equal. And the power rule, log(M^p) = p · log M, takes an exponent and moves it out front, where ordinary algebra can reach it.

The power rule applies to log(M^p) and nothing else. That's why isolating comes first: the log has to land on a bare power, not on a sum or a product that contains one.

Example 1: rewrite with a common base

Solve 9^{x + 1} = 27^x.

1. (3²)^{x + 1} = (3³)^x9 and 27 are both powers of 3. Write them that way.
2. 3^{2x + 2} = 3^3xPower of a power: multiply exponents. The 2 multiplies the whole x + 1.
3. 2x + 2 = 3xSame base on both sides, so the exponents must be equal.
4. x = 2Check: 9³ = 729 and 27² = 729. ✓

Example 2: isolate first

Solve 3 · 2^x = 48.

1. 2^x = 16Divide both sides by 3. The 3 multiplies the power; it is not part of the base.
2. 2^x = 2⁴Now 16 is visibly a power of 2.
3. x = 4Check: 3 · 2⁴ = 3 · 16 = 48. ✓

If you skip line 1 and try to make 3 · 2^x into a single power, you'll be tempted to write 6^x. That's wrong: the exponent belongs to the 2 only.

Example 3: bases that don't match, so take a log

Solve 4 · 3^x − 7 = 13.

1. 4 · 3^x = 20Add 7 to both sides.
2. 3^x = 5Divide by 4. The power is now alone.
3. ln(3^x) = ln 55 isn't a whole-number power of 3, so take ln of both sides. (log base 10 works just as well.)
4. x · ln 3 = ln 5Power rule: the exponent x comes down in front.
5. x = ln 5ln 3ln 3 is just a number (about 1.0986), so divide by it like any coefficient.
6. x ≈ 1.465Round only now. Check: 3^1.465 ≈ 5.00, and 4 · 5 − 7 = 13. ✓

ln 5 / ln 3 is the exact answer. It's also log₃ 5, by the change-of-base formula, which is a good way to read it: "the power of 3 that gives 5."

Example 4: x in both exponents

Solve 2^{x + 1} = 5^x.

1. ln(2^{x + 1}) = ln(5^x)2 and 5 have no common base. Take ln of both sides.
2. (x + 1) ln 2 = x ln 5Power rule on each side. Keep x + 1 in parentheses: the whole exponent comes down.
3. x ln 2 + ln 2 = x ln 5Distribute ln 2.
4. ln 2 = x ln 5 − x ln 2Collect the x terms on one side, like any linear equation.
5. ln 2 = x(ln 5 − ln 2)Factor out x.
6. x = ln 2ln 5 − ln 2 ≈ 0.756Divide. Check: 2^1.756 ≈ 3.38 and 5^0.756 ≈ 3.38. ✓

Once the logs are taken, ln 2 and ln 5 are constants, and this is a linear equation in x. The denominator can also be written ln(5/2), using the quotient rule.

Example 5: a quadratic in disguise

Solve 4^x − 3 · 2^x − 4 = 0.

1. (2^x)² − 3 · 2^x − 4 = 04^x = (2²)^x = (2^x)². Now every term is about 2^x.
2. u² − 3u − 4 = 0Let u = 2^x. It's an ordinary quadratic.
3. (u − 4)(u + 1) = 0−4 and 1 multiply to −4 and add to −3.
4. 2^x = 4 or 2^x = −1Put 2^x back in for u. These are values of u, not of x.
5. x = 22^x = 4 gives x = 2. 2^x = −1 has no solution: a positive base to any power is positive. Check: 16 − 12 − 4 = 0. ✓

Common mistakes, and the exact line they happen on

1. Taking the log before isolating

1. 2^x + 5 = 21Solve.
2. log(2^x) + log 5 = log 21This is the mistake. The log of a sum is not the sum of the logs. Following this line through gives x = log 4.2 / log 2 ≈ 2.07, and 2^2.07 + 5 ≈ 9.2, not 21.

Subtract 5 first: 2^x = 16, so x = 4. No log needed at all.

2. Treating the coefficient as part of the power

1. log(3 · 2^x) = log 48Logging 3 · 2^x = 48 without dividing first.
2. 3 · x log 2 = log 48This is the mistake. The power rule works on log(2^x), not log(3 · 2^x). The 3 is a separate factor: the product rule gives log 3 + x log 2. The wrong line leads to x ≈ 1.86, and 3 · 2^1.86 ≈ 10.9, not 48.

Done correctly, log 3 + x log 2 = log 48 gives x = (log 48 − log 3)/log 2 = log 16 / log 2 = 4. Dividing by 3 before taking any log gets there with less work.

3. Turning a quotient of logs into a log of a quotient

1. x = ln 5ln 3Correct so far, from Example 3.
2. x = ln 53 ≈ 0.511This is the mistake. The quotient rule is about a fraction inside one log. ln 5 / ln 3 is one log divided by another, and it equals about 1.465. Check the wrong value: 3^0.511 ≈ 1.75, not 5.

Type it into a calculator as ln(5) ÷ ln(3), with each log closed off before the division.

4. Rounding the logs too early

1. 1.1^x = 2How many years for money to double at 10% growth.
2. x = ln 2ln 1.1Take ln of both sides and divide. Exact so far.
3. x ≈ 0.690.1 = 6.9This is the mistake. ln 1.1 is 0.0953..., and rounding it to 0.1 changes it by about 5%. That error goes straight into the answer. The real value is ln 2 / ln 1.1 ≈ 7.27, and 1.1^6.9 ≈ 1.93, short of 2.

Keep the exact form ln 2 / ln 1.1 until the last line, then do the whole division on the calculator in one go.

5. Reporting the values of u as the values of x

1. (u − 4)(u + 1) = 0From Example 5, with u = 2^x.
2. x = 4 or x = −1This is the mistake. Those are values of u = 2^x, not x. Neither works: x = 4 gives 256 − 48 − 4 = 204, and x = −1 gives 1/4 − 3/2 − 4 = −5.25. Substitute back: 2^x = 4 gives x = 2, and 2^x = −1 is impossible, so it's rejected.

Practice

1. Solve 8^x = 14.

   Show answer

   Write both sides as powers of 2: 2^3x = 2⁻², so 3x = −2 and x = −2/3.

2. Solve 5^{2x − 1} = 125.

   Show answer

   125 = 5³, so 2x − 1 = 3 and x = 2.

3. Solve 2 · 7^x = 30. Give the exact answer and a decimal to three places.

   Show answer

   Isolate: 7^x = 15. Take ln: x ln 7 = ln 15, so x = ln 15 / ln 7 ≈ 1.392.

4. Solve 3^{x − 1} = 2^x.

   Show answer

   (x − 1) ln 3 = x ln 2, so x ln 3 − x ln 2 = ln 3 and x = ln 3 / (ln 3 − ln 2) ≈ 2.710.

5. Solve e^2x − e^x − 6 = 0.

   Show answer

   Let u = e^x: u² − u − 6 = 0, (u − 3)(u + 2) = 0. e^x = −2 is impossible, so e^x = 3 and x = ln 3 ≈ 1.099.