Problem:
$f(t) = 0.75 * 10^{kt}$ where k is a constant. We know that $f(2)=3$. Find $f(3)$.
My approach:
Rewrite $f(2)$ as: $0.75*10^{2k}=3$
Try to solve for k, eliminate 0.75 first: $10^{2k}=4$
This is where I get stuck: $log_{10}4=2k$
Where do I go from there? Am I doing something wrong? The answer is supposed to be $f(3)=6~$
Edit: typo in the logs, fixed. And I solved it, thanks everyone!
The statement $10^{2k}=4$ doesn't imply $2k=\log_4(10)$; it implies $2k=\log_{10}(4)$. So, you have that $$ k=\frac{\log_{10}(4)}{2}, $$ and therefore $$ f(3)=\frac{3}{4}\cdot 10^{3\log_{10}(4)/2}. $$ Your task now is to use the laws of logarithms/exponentials to simplify this expression.