For y=Abx, A>0,b>0, lg y=(lg b)x+lg A. For y=A ekx, ln y=kx+ln A. Taking logs makes the changing exponent a linear term. Plot log y against the original x. Exponentiate the intercept for A, and the gradient for b in the same base. With natural logs the gradient directly gives k in the base-e form. If ln y=ln8-xln2, then y=8(1/2)x. Its value halves whenever x increases by 1. Check: a negative logarithmic gradient represents decay, not a negative exponential base. Original y must be positive.

For \(y=Ab^x\), \(A>0,b>0\), \(\lg y=(\lg b)x+\lg A\). For \(y=A\mathrm e^{kx}\), \(\ln y=kx+\ln A\). Taking logs makes the changing exponent a linear term.

Plot log \(y\) against the original \(x\). Exponentiate the intercept for \(A\), and the gradient for \(b\) in the same base. With natural logs the gradient directly gives \(k\) in the base-e form.

If \(\ln y=\ln8-x\ln2\), then \(y=8(1/2)^x\). Its value halves whenever \(x\) increases by 1.

Check: a negative logarithmic gradient represents decay, not a negative exponential base. Original \(y\) must be positive.

Exponential models on logarithmic axes

Exponential models on logarithmic axes