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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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    Please write your question coherently.2011-03-14
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    @PEV, maybe he's using google translator. If I had the rep, I'd change "be" to "let"... (edit: thanks Henry!)2011-03-14
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    @The Chaz: people seem to be able to propose edits even if they don't have the rep to implement them. I don't know how, but I see the proposed edits (and have accepted a bunch and never rejected one).2011-03-14
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    @Ross *Now* I have the option to edit it. Maybe it's time- (or post-count) sensitive??2011-03-14
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    @The Chaz: I dunno. I know the ability to edit is rep sensitive, but I don't know how the low-rep people propose them. Maybe there is a delay to let the higher-rep people fix things first (which I would not support, but what do I know?) All I can suggest is to work with what you have-it sounds like you will be a positive contributor to the site.2011-03-14
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    @Ross Thanks! I'm more of an edit-the-imperative-out-of-the-question than a lecture-them-silly kind of guy, and I appreciate the occasional link to Lmgtfy.com - so I guess that only puts me at odds with ... one person?? :)2011-03-14
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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