2log25 - 3log5 + log20 =
2 log (5*5) -3 log 5 + log (2*2*5)
We know that: log (a*b)= log a + log b:
2 (log 5 + log 5) -3 log 5 + log 2 + log 2 + log 5=
4 log 5 - 3 log 5 +log 5 + 2 log 2=
2 log 5 + 2 log 2
You can take it just a little further Anonymous:
2log(5) + 2log(2) = log(52) + log(22) = log(25) + log(4) = log(25*4) = log(100) = 2 (assuming log to the base ten)
Great work anon.
I'm just going to look at it a slightly different way.
$$\\2log25 - 3log5 + log20 \\
=2log5^2-3log5+log4+log5\\
=4log5-3log5+log2^2+log5\\
=2log5+2log2\\
=log25+log4\\
=log100\\
=log10^2\\
=2log10\\
=2*1\\
=2$$
You beat me Alan.