Answer: \(y=3\times5^x\), so \(A=3\) and \(b=5\).

If \(\ln y\) plotted against \(x\) gives a straight line, write

\(\ln y=mx+c.\)

The line passes through \((1,\ln15)\) and \((2,\ln75)\), so its gradient is

\(m=\dfrac{\ln75-\ln15}{2-1}.\)

Using the logarithm law \(\ln A-\ln B=\ln\left(\dfrac AB\right)\),

\(m=\ln\left(\dfrac{75}{15}\right)=\ln5.\)

Now use the point \((1,\ln15)\):

\(\ln15=\ln5+c.\)

Therefore

\(c=\ln15-\ln5=\ln3.\)

So

\(\ln y=x\ln5+ \ln3.\)

Exponentiating both sides gives

\(y=\mathrm e^{x\ln5+\ln3}=\mathrm e^{\ln3}\mathrm e^{x\ln5}.\)

Since \(\mathrm e^{\ln3}=3\) and \(\mathrm e^{x\ln5}=5^x\),

\(y=3\times5^x.\)

Hence

\(A=3\quad\text{and}\quad b=5.\)