Can someone simplify this ($\log$ here refers to the common logarithm)? $$\sqrt{4\log2+(\log5)^2} + \sqrt{4\log5+(\log2)^2}$$ I know this has a simple solution but I cannot find it.
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Equations have solutions. Expressions do not. What you wrote is an expression and has no solution (the concept of a solution does not exist for it). Your question is like saying "What is the solution of $\pi$?" – 5xum Mar 20 '14 at 08:42
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Do you mean ".... $= 0$"? Or do you mean to simplify the expression? – Yiyuan Lee Mar 20 '14 at 08:43
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Because that would make a lot more sense. – user11977 Mar 20 '14 at 08:43
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yes, I mean to simplify and find what it is equal to – Anahit Mar 20 '14 at 08:44
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Do you have reason to believe this gets any simpler than it already is? – Mike Mar 20 '14 at 09:01
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yes, I even know that it is equal to 3 :D (I have the answer key) I just cannot understand how – Anahit Mar 20 '14 at 09:03
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Apparently WolframAlpha disagrees with the answer of $3$ : http://www.wolframalpha.com/input/?i=%5Csqrt%7B4%5Clog2%2B%28%5Clog5%29%28%5Clog5%29%7D+%2B+%5Csqrt%7B4%5Clog5%2B%28%5Clog2%29%28%5Clog2%29%7D, if you took $\log$ of base 10 instead of the natural $\log$. – Yiyuan Lee Mar 20 '14 at 09:12
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@YiyuanLee click the "use the base 10 logarithm instead" option. – Mike Mar 20 '14 at 09:16
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It seems pretty sneaky. So I'm guessing those are common logs. Let's take a look at the first square root. We have
$$\sqrt{4\log2+(\log5)^2}=\sqrt{4(\log 10-\log5)+(\log5)^2}=\sqrt{4-4\log5+(\log5)^2}=2-\log 5$$
Similarly, the second square root yields $2-\log 2$. Sum them together to get $4-\log10=3$.
Mike
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good catch on the common log. I was puzzling how one could simplify that at all using the natural log. On a math site, I assume natural unless labelled otherwise. Sneaky, indeed! (+1) – robjohn Mar 20 '14 at 17:59