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