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

2 Answers

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Given a^2+b^2=47ab. a>0 and b>0. To prove that log(a+b/7) = (1/2){loga+logb}.


a^2+b^2= 47ab.

We add 2ab to both sides:

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

=> (a+b)^2 = 49 ab.

We take the square root of both sides.

=> a+b = 7 *(ab)^(1/2).

Therefore (a+b)/7 = (ab)^(1/2).

We take logarithms of both sides:

loh{(a+b)/7} = log(ab)^(1/2)

log{(a+b)/7} = (1/2)log(ab), as log a^m = mloga.

log{(a+b)/7} =(1/2){loga+lob} , as log(mn) = logm+logn.