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Can't seem to figure out how to tackle this one. I know $\log 5 = 1 - \log 2$, but I don't see a way to get around the cubed logarithms except for brute force. The answer is $1$. Using the sum of cubes formula gets me $$ (\log 2 + 1 - \log 2)[(\log 2)^2 - \log 2(1 - \log 2) + (1 - \log 2)^2] \;, $$ which is so complex that I feel the question isn't meant to be solved this way. Any input would be appreciated. :)

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  • $\begingroup$ What base is $\log$? Natural logarithm? Logarithm to base 10? $\log 5 = 1-\log 2$ is only true for base 10. $\endgroup$ Dec 11, 2018 at 14:31
  • $\begingroup$ Sorry. I assumed that when the base is omitted it means base 10. $\endgroup$ Dec 11, 2018 at 14:37
  • $\begingroup$ @NamelessKing Look carefully. This expression is equal to $(\log2+\log5)^3$ $\endgroup$ Dec 11, 2018 at 15:02

4 Answers 4

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$(\log2)^3+(\log5)^3+(\log2)(\log125)\\=(\log2)^3+(\log5)^3+3(\log2)(\log5)(\log2+\log5)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \because \log2+\log5=\log10=1\\=(\log2+\log5)^3=1$

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Using your observation that $\log 5 = 1-\log 2$, along with $125 = 5^3$, we get $$ (\log 2)^3+(\log 5)^3+(\log 2)(\log 125)\\ = (\log 2)^3 + (1-\log 2)^3 + \log2\cdot 3\log 5\\ = (\log 2)^3 + 1 - 3\log 2 + 3(\log2)^2 - (\log 2)^3 + 3\log2(1-\log 2) $$ and we are soon left with just $1$.

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To simplify things let $x=\log(2)$ so that $\log(5)=1-x$. Also $$\log(125)=\log(5^3)=3\log(5)=3(1-x)$$ Now we get \begin{eqnarray*} (\log(2))^3+(\log(5))^3+\log(2)\log(125) & = & x^3+(1-x)^3+3x(1-x)\\ & = & x^3+1-3x+3x^2-x^3+3x-3x^2 \\ & = & 1 \end{eqnarray*}

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Hint: Expand $1 = (\log 2 + \log 5)^3$. Assuming base 10 here. Also note $125 = 5^3$.

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