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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?

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    @The Chaz: I suspect the link to Lmgtfy.com will put you at odds with more than one, but I am not included in that set. I tend to provide the link, but I agree a Google and Wikipedia search should be a prerequisite to posting. After all, posters are asking for free help from unknown strangers. So I feel that each of us should provide whatever help we feel like (based on how the question is asked, as well as what the question is, and our mood of the moment). And if you ask for help, you take the risk of not liking what you get. One view.2011-03-14

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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.

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    I would give +1 for the first line, even if it weren't right.2011-03-14
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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.

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    Yes, thank you. Fixed.2011-03-14