[SOLVED] How to change the base of y=A(1.3)^-t

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.

Changing exponential bases

Taking the last equality on your second line:

Quote:

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

You can simplify like this:

$\displaystyle y=e^{ln A + ln (1.3^{-t})}=e^{lnA} * e^{-t*ln1.3}$

$\displaystyle y=A*e^{-t*ln1.3}$

Another way of thinking of it is to solve the following for *x*:

$\displaystyle e^{x}=1.3$ to get $\displaystyle x=ln{1.3}$.

Then, instead of using 1.3 as your base, use $\displaystyle e^{ln{1.3}}$

It may seem strange to leave your answer in this form, without simplifying the *e* to the "*ln*" power, but there could be a variety of reasons for it. For example, now you know that $\displaystyle ln{1.3}$ is the rate of decay in a continuous decay model.