#1**+13 **

Best Answer

$$\\log 2ab + log 5/a - log10b = 0\\\\

LHS\\

=log\left(\frac{2ab\times \frac{5}{a}}{10b}\right)\\\\

=log\left(\frac{10b}{10b}\right)\\

=log\left(1\right)\\

=0\\

=RHS$$

Melody Aug 27, 2014

#2**+10 **

A property of logarithms is that log(x*y) = log(x) + log(y)

Another is that log(x^{-1}) = -log(x)

So: log(2*a*b) = log(2) + log(a) + log(b)

log(5/a) = log(5) + log(1/a) = log(5) - log(a) (as 1/a = a^{-1})

log(10*b) = log(10) + log(b)

Therefore: log(2*a*b) + log(5/a) - log(10*b) = log(2) + log(a) + log(b) + log(5) - log(a) - log(10) - log(b)

= log(2) + log(5) - log(10)

= log(10) - log(10)

= 0

I see Melody beat me to it (with a slightly different approach)!

Alan Aug 27, 2014