Be aware of Logs!

Given

log ( a 2 + b 2 ) = 1 2 ( log a + log b ) \log \left(\frac{a}{2} + \frac{b}{2}\right) = \frac{1}{2}(\log a + \log b)

Find the value of a b a - b .


The answer is 0.

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2 solutions

Antony Diaz
Jun 4, 2014

log a + b 2 = 1 2 log a b 2 log a + b 2 = log a b log ( a + b 2 ) 2 = log a b \log { \frac { a+b }{ 2 } } =\frac { 1 }{ 2 } \log { ab } \\ 2\log { \frac { a+b }{ 2 } } =\log { ab } \\ \log { { (\frac { a+b }{ 2 } ) }^{ 2 } } =\log { ab }

Now, we can equate: ( a + b ) 2 4 = a b a 2 + 2 a b + b 2 = 4 a b ( a b ) 2 = 0 \frac { { (a+b) }^{ 2 } }{ 4 } =ab\\ { a }^{ 2 }+2ab+{ b }^{ 2 }=4ab\\ { (a-b) }^{ 2 }=0

And we obtain that a b = 0 a-b=0

Kartik Sharma
Jun 4, 2014

log (a+b)/2 = 1/2 (log ab)

[log (a+b)/2]/log ab = 1/2

log to the base ab (a+b/2) = 1/2

Hence, (a+b)/2 = (ab)^1/2

a+b = 2(ab)^1/2

Squaring both sides,

(a + b)^2 = 4 ab

a^2 + b^2 = 2ab

So, (a-b)^2 = 2ab-2ab = 0

Therefore, a-b = 0

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