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# Why does log 2ab + log 5/a - log10b = 0 ?

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Why does log 2ab + log 5/a - log10b = 0 ?

math
Aug 27, 2014

#1
+13

$$\\log 2ab + log 5/a - log10b = 0\\\\ LHS\\ =log\left(\frac{2ab\times \frac{5}{a}}{10b}\right)\\\\ =log\left(\frac{10b}{10b}\right)\\ =log\left(1\right)\\ =0\\ =RHS$$

.
Aug 27, 2014

#1
+13

$$\\log 2ab + log 5/a - log10b = 0\\\\ LHS\\ =log\left(\frac{2ab\times \frac{5}{a}}{10b}\right)\\\\ =log\left(\frac{10b}{10b}\right)\\ =log\left(1\right)\\ =0\\ =RHS$$

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)!

Aug 27, 2014