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

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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 yields:

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

Reducing duplicate factors yields:

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

You need to convert the square root into a rational power, such that:

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

Using logarithmic identities yields:

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

You need to convert the logarithm of product a*b into a summation of logarithms, such that:

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

Hence, testing the validity of expression `log ((a + b)/7) = 1/2*(log a + log b)` yields that it holds under the given conditions.

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.

We’ve answered 318,908 questions. We can answer yours, too.

Ask a question