Change-of-Base Formula

The change-of-base formula converts a logarithm in any base to a ratio of logarithms in a convenient base: log_b x = log_c x / log_c b. In brief: to find log₂ 50 on a calculator that has only log and ln, compute ln 50/ln 2 = 3.912/0.6931 ≈ 5.644. Special cases: log_b a = 1/log_a b, and log_b^k x = (1/k) log_b x.

Calculators typically offer only two logarithm keys, for base 10 and for base e. The formula lets you evaluate any other base and also helps in simplifying expressions with different bases and in solving exponential equations. This guide states and proves the formula, shows how to use it with a calculator, derives useful consequences and gives a table of conversions between bases.

What is the change-of-base formula?

For b, c > 0 with b ≠ 1 and c ≠ 1, and x > 0:

log_b x = log_c x / log_c b.

The base c can be any valid base. The most common choices are c = 10 and c = e: log_b x = log x/log b = ln x/ln b.

Why does the formula hold?

Let y = log_b x, so b^y = x. Take the logarithm in base c of both sides: log_c(b^y) = log_c x. By the power rule, y log_c b = log_c x. Dividing by log_c b (which is not zero since b ≠ 1) gives y = log_c x/log_c b. ∎

How do you use it with a calculator?

1. Choose the base of your calculator’s key: log (base 10) or ln (base e).
2. Compute the logarithm of the argument.
3. Compute the logarithm of the base.
4. Divide the first result by the second.

Example 1: log₂ 50

Using ln: ln 50 = 3.9120 and ln 2 = 0.6931, so log₂ 50 = 5.6439. Using log: log 50 = 1.6990 and log 2 = 0.3010, so 1.6990/0.3010 = 5.644. The same result, as expected. Check: 2^5.644 ≈ 50.

Example 2: log₇ 100

ln 100/ln 7 = 4.6052/1.9459 = 2.3667. Check: 7^2.3667 ≈ 100, since 7² = 49 and 7³ = 343.

Example 3: a base less than 1

log_0.5 10 = ln 10/ln 0.5 = 2.3026/(−0.6931) = −3.3219.

What are the useful consequences?

• Reciprocal bases: log_b a = 1/log_a b. (Set x = a and c = a in the formula.) For example, log₂ 8 = 3 and log₈ 2 = 1/3.
• Power of the base: log_b^k x = (1/k) log_b x. For instance, log₉ 27 = (1/2) log₃ 27 = 3/2.
• Power of the argument: log_b^k (x^m) = (m/k) log_b x.
• Chain rule for bases: log_a b · log_b c = log_a c.
• Swapping: a^{log_b c} = c^{log_b a}.

Worked examples

Example 4: simplifying without a calculator

log₄ 32 = log₂ 32/log₂ 4 = 5/2. And log₈ 16 = log₂ 16/log₂ 8 = 4/3.

Example 5: the chain rule for bases

log₂ 3 · log₃ 4 · log₄ 5 · log₅ 8 = log₂ 8 = 3. Each factor cancels its neighbour’s base.

Example 6: solving an exponential equation

5^x = 37. Then x = log₅ 37 = ln 37/ln 5 = 3.6109/1.6094 = 2.2436. Check: 5^2.2436 ≈ 37.0.

Example 7: comparing numbers

Which is larger, log₃ 7 or log₅ 20? ln 7/ln 3 = 1.9459/1.0986 = 1.7712, and ln 20/ln 5 = 2.9957/1.6094 = 1.8614. So log₅ 20 is larger.

Example 8: equation with two bases

log₂ x + log₄ x = 3. Convert: log₄ x = (1/2) log₂ x. Then (3/2) log₂ x = 3, so log₂ x = 2 and x = 4. Check: 2 + 1 = 3 ✓.

Example 9: doubling time

A population growing 3% per year doubles after t = log_1.03 2 = ln 2/ln 1.03 = 0.6931/0.02956 ≈ 23.45 years.

Conversion table between bases

For example, log₂ x = 1.4427 ln x, and log x = 0.4343 ln x. The factor 1/ln 2 ≈ 1.4427 converts natural logarithms to base-2 logarithms.

Where does the formula come from?

When tables of logarithms were the standard computational tool, the change-of-base formula allowed one table, for example base 10, to serve all other bases. Napier’s original logarithms were not of a standard base, and Briggs’s common logarithms in base 10 became the standard because of their link with the decimal system. The formula expresses the fact that logarithm functions in different bases are proportional to each other.

Where is it used today?

Computer scientists convert between log₂ and ln to analyse algorithms: the number of bits needed for n values is ⌈log₂ n⌉ = ⌈ln n/ln 2⌉. In chemistry and acoustics, the formula converts between the logarithmic units. Software libraries implement log₂ and log₁₀ directly or by this formula.

Which mistakes are common?

Inverting the ratio

log_b x = log x/log b, not log b/log x. Check with log₂ 8 = 3: log 8/log 2 = 0.9031/0.3010 = 3 ✓.

Cancelling the logs

log x/log b is not x/b. The logarithm function cannot be cancelled.

Using different bases in the numerator and denominator

Both logarithms must use the same new base c.

Rounding too early

Use at least five decimals in intermediate steps. A rounded value of ln 2 as 0.69 can give an error in the second decimal place.