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

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

The question says:
" $y=A(1.3)^{-t}$
By changing the base of the exponent to e, express y in terms of A, e and t."
I tried:
$y=A(1.3)^{-t}\Rightarrow\frac{y}{A}=1.3^{-t} \Rightarrow log_{1.3}\frac{y}{A}=-t\Rightarrow\frac{\ln\frac{y}{A}}{\ln1.3}=-t$

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

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

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

2. ## Changing exponential bases

Taking the last equality on your second line:
$\Rightarrow\ln\frac{y}{A}=-t\ln 1.3\Rightarrow\ln y-\ln A=\ln1.3^{-t}\Rightarrow\ln y=\ln A+\ln1.3^{-t}$
You can simplify like this:
$y=e^{ln A + ln (1.3^{-t})}=e^{lnA} * e^{-t*ln1.3}$
$y=A*e^{-t*ln1.3}$

Another way of thinking of it is to solve the following for x:
$e^{x}=1.3$ to get $x=ln{1.3}$.
Then, instead of using 1.3 as your base, use $e^{ln{1.3}}$

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 $ln{1.3}$ is the rate of decay in a continuous decay model.

3. Thanks a lot, that's a really great answer. : )