Ok great thanks
$\displaystyle log_bA = 2$
$\displaystyle log_bD = 5$
I presume you want to know: $\displaystyle log_b(A+D)$? (Yes, case is important. A is not necessarily the same as a.)
I couldn't tell you. With the given information I could tell you what $\displaystyle log_b(AD)$ is (its 2 + 5 = 7).
Let me show you why this is so hard.
$\displaystyle log_bA = 2$ means that $\displaystyle b^2 = A$. Similarly $\displaystyle log_bD = 5$ means $\displaystyle b^5 = D$. So
$\displaystyle A + D = b^2 + b^5$
But this is NOT a simple power of b. There IS an exponent c such that $\displaystyle b^c = b^2 + b^5$, but this is not an easy equation to solve for a general value of b, if it's even possible to do generally. (However if we know the value of b we can numerically estimate it.)
If it still seems simple, try to find c for $\displaystyle 2^c = 2^2 + 2^5 = 4 + 32 = 36$. So $\displaystyle c = log_236 \approx 5.169925001$. But if b = 3, then $\displaystyle c = log_2252 \approx 5.033103256$.
-Dan
Dan you are right I did A time D and got 15 and that was wrong on the test. Maybe I am in way over my head, I might should have just added them. But when you expand logs through mulitiplication you add them, this is what I thought but as stated she marked it wrong. My teacher obviously wanted something totally different. Thanks Guys.