Question

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show that :-

If a > 1, then a² < a.

#Advanced Mathematics!

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Answer 1

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Remember that:

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

and

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

So,

a2+b2=7aba2+b2=7ab

Adding 2ab2ab on both sides :

a2+b2+2ab=7ab+2aba2+b2+2ab=7ab+2ab

(a+b)2=9ab(a+b)2=9ab

a+b=9ab−−−√=9–√⋅ab−−√=3ab−−√a+b=9ab=9⋅ab=3ab

a+b=3ab−−√a+b=3ab

Multiply by 1313 on both sides :

13(a+b)=13⋅3ab−−√13(a+b)=13⋅3ab

13(a+b)=ab−−√=(ab)1213(a+b)=ab=(ab)12

13(a+b)=(ab)1213(a+b)=(ab)12

Log on both sides :

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

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

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