Logarithmic and Exponential Functions · 0606 Topic 6

Change of Base

Rig, founder of IGCSE Add Math Malaysia

Written by Rig, our founder

8 years teaching IGCSE & SPM maths · Updated 16 August 2026

The change-of-base rule lets you rewrite a logarithm to any base logax\log_a x in terms of logs your calculator has (ln\ln or lg\lg). It’s what makes exponentials with awkward bases solvable.

The rule

logax=logbxlogba\log_a x = \frac{\log_b x}{\log_b a}

In practice, use ln\ln (or lg\lg), since those are the calculator buttons: logax=lnxlna\log_a x = \frac{\ln x}{\ln a}

So log27=ln7ln22.81\log_2 7 = \dfrac{\ln 7}{\ln 2} \approx 2.81.

Solving exponentials with it

The rule is really the same move as solving bx=cb^x = c with logs. To solve 2x=72^x = 7, take logs and divide: x=log27=ln7ln22.81x = \log_2 7 = \frac{\ln 7}{\ln 2} \approx 2.81

Whether you call it “change of base” or “take logs and use the power law”, the arithmetic is identical: a division of two logs.

The key warning

ln7ln2\dfrac{\ln 7}{\ln 2} is a division, not ln7ln2\ln 7 - \ln 2. The subtraction law would give ln72\ln\frac{7}{2}, a completely different value. This mix-up is the most common error with change of base: ln7ln22.81butln7ln2=ln3.51.25\frac{\ln 7}{\ln 2} \approx 2.81 \qquad \text{but} \qquad \ln 7 - \ln 2 = \ln 3.5 \approx 1.25

They’re not the same. The change-of-base ratio divides the logs; it does not subtract them.

Why the base doesn’t matter

You can use ln\ln or lg\lg (or any base) in the ratio and get the same answer, because the base cancels. ln7ln2=lg7lg2\dfrac{\ln 7}{\ln 2} = \dfrac{\lg 7}{\lg 2}. Use whichever your calculator offers.

Common mistakes

  • Subtracting the logs instead of dividing them.
  • Inverting the ratio (ln2ln7\frac{\ln 2}{\ln 7}).
  • Rounding the two logs before dividing, losing accuracy.

Full topic context: Logs & Exponentials notes, and Laws of Logarithms.

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