Question

If a²-11ab+b²=0, then let us show that log ⅓(a-b)=½ (log a + log b).​

Answer 1

Step-by-step explanation:

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))