For a>0, a 1 and b>0, au=b gives uln a=ln b. Thus u= ln bln a. This is change of base applied to an equation. Method: isolate the exponential expression; check that the other side is positive; take logarithms; use the power law; solve the resulting linear equation. Keep the exact logarithmic form until the last step. For 3 e2x-1=11, first e2x=4, then 2x=ln4, so x=ln2. With unlike bases on both sides, take logarithms of the whole of each side: 2x=3x-1 gives xln2=(x-1)ln3. Exam check: a positive exponential never reaches zero. In a time model, reject a time outside the stated interval even when the algebra gives a real number. Round only the final value.

For \(a>0\), \(a\ne1\) and \(b>0\), \(a^u=b\) gives \(u\ln a=\ln b\). Thus \(u=\dfrac{\ln b}{\ln a}\). This is change of base applied to an equation.

Method: isolate the exponential expression; check that the other side is positive; take logarithms; use the power law; solve the resulting linear equation. Keep the exact logarithmic form until the last step.

For \(3\mathrm e^{2x}-1=11\), first \(\mathrm e^{2x}=4\), then \(2x=\ln4\), so \(x=\ln2\).

With unlike bases on both sides, take logarithms of the whole of each side: \(2^x=3^{x-1}\) gives \(x\ln2=(x-1)\ln3\).

Exam check: a positive exponential never reaches zero. In a time model, reject a time outside the stated interval even when the algebra gives a real number. Round only the final value.

Taking logarithms to find an exponent

Taking logarithms to find an exponent