math If a^2+b^2=47ab where a>0 and b>0 show that log(a+b/7)=1/2(loga+logb).

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You also may start by completing the square in the given relation `a^2 + b^2 = 47 ab` , such that:

`a^2 + 2ab + b^2 = 47 ab + 2ab`

`(a + b)^2 = 49 ab => a + b = sqrt(49 ab) => a + b = 7sqrt(ab)`

Replacing `7sqrt(ab)` for` a + b` in expression you need to check out...

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giorgiana1976 | Student

We'll re-write the given expression, applying the product rule of logarithms:

log a + log b = log a*b

log (a+b)/7 = 1/2*log (ab)

We'll apply the power rule of logarithms:

a*log b = log b^a =>  1/2*log (ab) = log (ab)^(1/2)

But  (ab)^(1/2) = sqrt (ab)

log (a+b)/7 = log [sqrt (ab)]

Since the bases are matching, we'll apply one to one rule:

(a+b)/7 = sqrt (ab)

We'll multiply by 7 both sides:

a + b = 7sqrt (ab)

We'll raise to square both sides:

(a+b)^2 = 49ab

We'll expand the square:

a^2 + 2ab + b^2 = 49ab

We'll subtract 2ab both sides:

a^2 + b^2 = 49ab - 2ab

a^2 + b^2 = 47ab q.e.d.

We've get the constraint given by enunciation so, the expression

log (a+b)/7 = (1/2)*(log a + log b) is true for a^2 + b^2 = 47ab.

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