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

math
Guest Aug 27, 2014

Best Answer 

 #1
avatar+93627 
+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
 #1
avatar+93627 
+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
avatar+27035 
+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

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