Originally Posted by

**Jhevon** you actually ended up using the same formula without trealizing it. now you are correct so far. now all that remains is to find $\displaystyle t$. we need logs for this. and you should remember, whenever you see a variable as a power and you need to solve for the variable, chances are you need logs.

let's pick up where you left off.

$\displaystyle 800(1.08)^t = 1000(1.06)^t$ .....divide both sides by 1000

$\displaystyle \Rightarrow \frac {4}{5} (1.08)^t = 1.06^t$ ......take log base 10 of both sides

$\displaystyle \Rightarrow \log \left( \frac {4}{5} (1.08)^t \right) = \log 1.06^t$

$\displaystyle \Rightarrow \log \left( \frac {4}{5} \right) + \log (1.08)^t = \log 1.06^t$ ...since $\displaystyle \log_a xy = \log_a x + \log_a y$

$\displaystyle \Rightarrow \log (1.08)^t - \log (1.06)^t = - \log \frac {4}{5}$......group $\displaystyle t$'s on one side

$\displaystyle \Rightarrow \log \left( \frac {1.08}{1.06} \right)^t = - \log \frac {4}{5}$ ......since $\displaystyle \log_a \left( \frac {x}{y} \right) = \log_a x - \log_a y$

$\displaystyle \Rightarrow t \log \left( \frac {1.08}{1.06} \right) = - \log \frac {4}{5}$ ......since $\displaystyle \log_a x^n = n \log_a x$

$\displaystyle \Rightarrow t = - \frac { \log \frac {4}{5}}{ \log \left( \frac {1.08}{1.06}\right)}$

And now just plug that into your calculator

EDIT: Ah, CaptainBlack beat me to it, oh well