An arithmetic sequence of positive logarithms Suppose $a$, $b$, $c$, and $d$ are positive numbers and each is not equal to $1$. 
If  $\log_a(d)$, $\log_b(d)$, and $\log_c(d)$  are an arithmetic progression in this order, then what is $(ac)^{\log_a(b)}$ equal to?
 A: It is equal to a lot of things, but I suspect that $c^2$ is one of them.
You can show this by knowing that $\log_a(d) = \dfrac{\log_e(d)}{\log_e(a)}$ and similarly with other letter combinations.
Then the arithmetic progression implies a harmonic progression of the logarithms of $a$, $b$ and $c$.  That will then tell you about $\log_e(ac)$, which you can then multiply by $\log_e(b) / \log_e(a)$.  If you then tidy up the right hand side, you should get something like $2 \log_e(c)$.  And then it is one simple step to the solution.
A: I take this to mean that $\log_c(d) - \log_b(d) = \log_b(d) - \log_a(d) = x$ for some $x$. In this case, the key is to use the properties of logarithms (including addition and multiplication of logarithms and changing the base) to manipulate these into a useful expression. My approach would be to start with $(ac)^{\log_{a}b}$, which I simplified to $b*c^{\log_{a}b}$, and using the difference relation derive $\log_{a}b = b/(1/(\log_{b}c)-1+b)$. Assuming that I made no mistakes in manipulation for the last derivation, this should easily give you an answer.
