Question

If x^ + y^ = z^ show that log z- y base x +log z+y base x =2​

Answer 1

Explanation:

Let a=x ^logy−logz

logy−logz ,b=y ^logz−logx

logz−logx ,c=z ^logx−logy

logx−logy

logx−logy Now, log(abc)=loga+logb+logc

logx−logy Now, log(abc)=loga+logb+logc⇒log(abc)=(logy−logz)logx+(logz−logx)logy+(logx−logy)logz=0

logx−logy Now, log(abc)=loga+logb+logc⇒log(abc)=(logy−logz)logx+(logz−logx)logy+(logx−logy)logz=0⇒abc=1

logx−logy Now, log(abc)=loga+logb+logc⇒log(abc)=(logy−logz)logx+(logz−logx)logy+(logx−logy)logz=0⇒abc=1As a,b,c are positive AM≥GM

logx−logy Now, log(abc)=loga+logb+logc⇒log(abc)=(logy−logz)logx+(logz−logx)logy+(logx−logy)logz=0⇒abc=1As a,b,c are positive AM≥GM∴ 3/a+b+c≥(abc) 3/1 =1

1 =1∴a+b+c≥3

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