Question

ifa²+b²=7ab log(1/3(a+b))=1/2(loga+logb)​

Answer 1

Answer:

Remember that:

(a+b)2=a2+2ab+b2

and

log(a)+log(b)=log(ab)

So,

a2+b2=7ab

Adding 2ab on both sides :

a2+b2+2ab=7ab+2ab

(a+b)2=9ab

a+b=9ab−−−√=9–√⋅ab−−√=3ab−−√

a+b=3ab−−√

Multiply by 13 on both sides :

13(a+b)=13⋅3ab−−√

13(a+b)=ab−−√=(ab)12

13(a+b)=(ab)12

Log on both sides :

log(13(a+b))=log((ab)12)=12log(ab)

But 12log(ab)=12(log(a)+log(b)) . So,

log(13(a+b))=12(log(a)+log(b))

Answer 2

Given that,

$\sf { a}^{2} + b {}^{2} = 7ab$

Adding 2ab to both sides,

$\sf \longrightarrow { a}^{2} + b {}^{2} + 2ab = 9ab$

$\sf \longrightarrow {( a + b)}^{2} = {(3 \sqrt{ab}) }^{2}$

$\sf \longrightarrow a + b =3 \sqrt{ab} \quad \quad \dots(1)$

Now it is given to prove that,

$\sf \: log \bigg \{\dfrac{1}{3} (a + b) \bigg\} = \dfrac{1}{2} \bigg \{ log(a) + log(b) \bigg \}$

Considering LHS,

$\sf \: log \bigg \{\dfrac{1}{3} (a + b) \bigg\}$

Now from (1),

$\sf \longrightarrow log \bigg \{\dfrac{1}{3} \times 3 \sqrt{ab} \bigg\}$

$\sf \longrightarrow log( \sqrt{ab} )$

$\sf \longrightarrow log( {ab}^{1/2} )$

Now, we know that,

$\red{ \bigstar} \boxed{ \sf{ log( {y}^{x} ) = x log(y) }}$

So, we can write,

$\sf \longrightarrow \dfrac{1}{2} log( ab )$

Now, we know that,

$\orange{ \bigstar} \boxed{ \sf{ log( xy ) = log(x) + log(y) }}$

So, we can write,

$\sf \longrightarrow \dfrac{1}{2} \bigg \{ log( a ) + log(b) \bigg \}$

Considering RHS,

$\sf \dfrac{1}{2} \bigg \{ log(a) + log(b) \bigg \}$

Thus LHS = RHS.

Hence proved!