The question says:"$\displaystyle y=A(1.3)^{-t}$

By changing the base of the exponent to e, express y in terms of A, e and t."

I tried:

$\displaystyle y=A(1.3)^{-t}\Rightarrow\frac{y}{A}=1.3^{-t} \Rightarrow log_{1.3}\frac{y}{A}=-t\Rightarrow\frac{\ln\frac{y}{A}}{\ln1.3}=-t$

$\displaystyle \Rightarrow\ln\frac{y}{A}=-t\ln 1.3\Rightarrow\ln y-\ln A=\ln1.3^{-t}\Rightarrow\ln y=\ln A+\ln1.3^{-t}$

$\displaystyle \Rightarrow\ln y=\ln A(1.3)^{-t}\Rightarrow y=A(1.3)^{-t}$

It just goes round in a circle. What am I doing wrong?

Thanks in advance.