Question

i need help solving this for an alegebra project.
3600=(.0059)(2ⁿ)

Answer 1

solving for n
\[3600=2^n(0.0059) \] \[2^n={3600 \over 0.0059}\]take log base two of both sides:\[n=\log_2({3600 \over 0.0056})\approx \log_2(610169)\]my calculator can't do log base two, but you get the point I hope.

Answer 2

im still stuck! HELP

Answer 3

With what? The final line above is the answer. Please explain where you are stuck, or what it is that you don't understand.