Start with \(y=e^{ax+b}\).

Take natural logarithms of both sides:

\(\ln y=ax+b\).

Compare this with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=x,\ m=a,\ c=b\).

Start with \(y=10^{ax-b}\).

Take logarithms base \(10\) of both sides:

\(\log_{10} y=ax-b\).

Compare this with \(Y=mX+c\).

So a suitable choice is

\(Y=\log_{10} y,\ X=x,\ m=a,\ c=-b\).

Start with \(y=ax^{-b}\).

Take natural logarithms:

\(\ln y=\ln(a x^{-b})\).

Use log laws:

\(\ln y=\ln a+\ln(x^{-b})\)

\(\ln y=\ln a-b\ln x\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=\ln x,\ m=-b,\ c=\ln a\).

Start with \(y=ab^x\).

Take natural logarithms:

\(\ln y=\ln(ab^x)\).

Use log laws:

\(\ln y=\ln a+\ln(b^x)\)

\(\ln y=\ln a+x\ln b\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=\ln y,\ X=x,\ m=\ln b,\ c=\ln a\).

Start with \(a=e^{x^2+by}\).

Take natural logarithms:

\(\ln a=x^2+by\).

Rearrange to make \(y\) the subject:

\(by=-x^2+\ln a\)

\(y=-\dfrac{1}{b}x^2+\dfrac{\ln a}{b}\).

Compare with \(Y=mX+c\).

So a suitable choice is

\(Y=y,\ X=x^2,\ m=-\dfrac{1}{b},\ c=\dfrac{\ln a}{b}\).