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Logarithms and exponentials

log_b(x) in any base, a power b^y, or a^x = c solved, with the logarithm rules and their conditions of existence.

Parameters

10: the decimal log. 2: computing. 2.71828: an approximate ln.

Result

Logarithm

3

Calculation
log10(1000) = ln(1000) / ln(10) = 3
Natural logarithm (ln)
6.90776
Binary logarithm
9.96578
Notes (2)

You must raise 10 to the power 3 to obtain 1000.

log(a + b) does not simplify. Only products, quotients and powers expand: log(ab) = log a + log b, log(a/b) = log a − log b, log(aⁿ) = n·log a.

Calculation, assumptions and sources

OverviewLogarithm in the chosen base, with ln, decimal log and binary log; power b^y; solution of a^x = c by change of base.

Logarithm in the chosen base, with ln, decimal log and binary log; power b^y; solution of a^x = c by change of base.

Method & formulaslog_b(x) = ln x / ln b
logb(x) = ln x / ln ba^x = c ⇔ x = ln c / ln alog(ab) = log a + log blog(aⁿ) = n × log a

The change of base brings every logarithm back to the natural one and solves exponential equations. The logarithm turns products into sums and powers into products.

Logarithm of base b
the exponent that brings b to x: log_b(x) = y means b^y = x.
Natural logarithm (ln)
base e ≈ 2.71828; the derivative of ln x is 1/x.
Decimal logarithm (log₁₀)
one unit of log₁₀ is a factor of 10. Base of pH, decibels and magnitudes.
Binary logarithm (log₂)
base of information (bits): log₂(1024) = 10, ten doublings.
Validity domainThe calculation is in the reals: a strictly positive number, a base strictly positive and different from 1.

The calculation is in the reals: a strictly positive number, a base strictly positive and different from 1. A negative base with a non-integer exponent leaves the reals and is not covered. Entered numbers must stay between about 10⁻³⁰⁸ and 10³⁰⁸.

The logarithm of a sumlog(a + b) does not simplify and is not log a + log b: log₁₀(10 + 10) = log₁₀(20) ≈ 1.3, while log₁₀(10) + log₁₀(10) = 2.

log(a + b) does not simplify and is not log a + log b: log₁₀(10 + 10) = log₁₀(20) ≈ 1.3, while log₁₀(10) + log₁₀(10) = 2.

Reading the resultFor 1000, log₁₀ = 3 (three orders of magnitude), log₂ ≈ 9.97 (about 10 doublings) and ln ≈ 6.9.

For 1000, log₁₀ = 3 (three orders of magnitude), log₂ ≈ 9.97 (about 10 doublings) and ln ≈ 6.9. A negative logarithm corresponds to a number between 0 and 1: log₁₀(0.001) = −3. At 7% per period, 1.07^x = 2 gives a doubling time x ≈ 10.24 periods.