The value of $\log_{8}{2938}$ is between the consecutive integers $a$ and $b$.  Find $a+b$.
I know that $8^4 = 4096$, which is too big, so $\log_{8}{2938}$ must be less than 4. I also know that $8^3 = 512$, which is too small, so $\log_{8}{2938}$ must be greater than 3. Therefore, $\log_{8}{2938}$ is between 3 and 4, and $a = 3$ and $b = 4$. The answer is $a+b = 3+4 = 7$.

