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.
Best Answer
$\log16+\log25-\log36=\log(\frac{16\cdot25}{36})=\log((\frac{4\cdot5}{6})^2)=2\log(\frac{20}{6})=2\log(\frac{10}{3})$
$\log10-\log3=\log(\frac{10}{3})$
$\frac{2\log(\frac{10}{3})}{\log(\frac{10}{3})}=2$