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The problem I have is: $\log16+\log25-\log36\over{\log10-\log3}$

(log is base 10 here)

I have the answer as 2 but no idea how to reach it..

I need to work this out without the use of a calculator but I can't get my head round it. I know that $\log(16) = \log(4^2) = 2(\log4)$ $\log25 = 2(\log5) $

I can add \logs together when they all are a power of the same number e.g. $\log(64) + \log(32) = \log(2^6) + \log(2^5) = 6(\log2) + 5(\log2) = 11(\log2)$.

I might just be over thinking it or over complicating it but I'd really appreciate some help here. Thanks.

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    $\begingroup$ 2 is the answer :) Use $\log a + \log b - \log c = \log \frac{ab}{c}$ $\endgroup$ – Mann May 12 '15 at 7:01
  • $\begingroup$ Yeah I forgot to mention I know the answer, but no idea how to actually work it out. Thanks $\endgroup$ – Ordered May 12 '15 at 7:08
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$\log16+\log25-\log36=\log(\frac{16\cdot25}{36})=\log(\frac{20}{6})^2=2\log(\frac{20}{6})=2\log(\frac{10}{3})$

$\log10-\log3=\log(\frac{10}{3})$

$\frac{2\log(\frac{10}{3})}{\log(\frac{10}{3})}=2$

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Hint: Use $\log a +\log b=\log ab$ and $\log a -\log b=\log \frac{a}{b}$

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