equivalent equation to y=2.5(2) ^t/5 show work

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*2^t/5)

log₂y = log₂(2.5) + log₂(2^t/5)

log₂y = log₂(2.5) + t/5

t/5 = log₂y - log₂(2.5)

t/5 = log₂(y/2.5)

t = 5log₂(y/2.5)

Thanks for your help!