Alan's answer is excellent but I would not have thought to do it that way.
also I often refer to this page when I do log questions.
http://en.wikipedia.org/wiki/List_of_logarithmic_identities
Here is a more mainstream solution.
$$\begin{array}{rll}
y&=&2.5\times 2^{t/5}\\\\
\frac{y}{2.5}&=&2^{t/5}\\\\
log\left(\frac{y}{2.5}\right)&=&log(2^{t/5})\\\\
log\left(\frac{y}{2.5}\right)&=&(t/5)log(2)\\\\
\frac{log\left(\frac{y}{2.5})}{log(2)}\right)&=&t/5\\\\
\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)&=&t\\\\
t&=&\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)\\\\
t&=&\frac{5log\left(0.4y)}{log(2)}\right)\\\\
\end{array}$$
I'd stop here but i want to check it is the same as Alan's solution.
I know that
$$\frac{log_{10}A}{log_{10}B}=Log_BA$$
so
$$t=5log_2(y/2.5)\\
t=5log_2(0.4y)$$
They are the same.
I assume that by "equivalent equation" you mean rearrange to get t = ...
Take log to the base 2 of both sides
log2y = log2(2.5*2t/5)
log2y = log2(2.5) + log2(2t/5)
log2y = log2(2.5) + t/5
t/5 = log2y - log2(2.5)
t/5 = log2(y/2.5)
t = 5log2(y/2.5)
Alan's answer is excellent but I would not have thought to do it that way.
also I often refer to this page when I do log questions.
http://en.wikipedia.org/wiki/List_of_logarithmic_identities
Here is a more mainstream solution.
$$\begin{array}{rll}
y&=&2.5\times 2^{t/5}\\\\
\frac{y}{2.5}&=&2^{t/5}\\\\
log\left(\frac{y}{2.5}\right)&=&log(2^{t/5})\\\\
log\left(\frac{y}{2.5}\right)&=&(t/5)log(2)\\\\
\frac{log\left(\frac{y}{2.5})}{log(2)}\right)&=&t/5\\\\
\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)&=&t\\\\
t&=&\frac{5log\left(\frac{y}{2.5})}{log(2)}\right)\\\\
t&=&\frac{5log\left(0.4y)}{log(2)}\right)\\\\
\end{array}$$
I'd stop here but i want to check it is the same as Alan's solution.
I know that
$$\frac{log_{10}A}{log_{10}B}=Log_BA$$
so
$$t=5log_2(y/2.5)\\
t=5log_2(0.4y)$$
They are the same.