Evaluate the log expression \frac{1}{\log_{xy}(xyz)}+\frac{1}{\log_{yz}(xyz)}+\frac{1}{\log_{zx}(xyz)}

Annette Sabin

Annette Sabin

Answered question

2022-01-20

Evaluate the log expression
1logxy(xyz)+1logyz(xyz)+1logzx(xyz)

Answer & Explanation

poleglit3

poleglit3

Beginner2022-01-20Added 32 answers

Use the fact that
logab=lnblna
Thus
1logxy(xyz)=ln(xy)ln(xyz)
ect.
Mary Goodson

Mary Goodson

Beginner2022-01-21Added 37 answers

By using the identity
logb(a)=lnalnb
Your equation becomes:
ln(xy)ln(xyz)+ln(xz)ln(xyz)+ln(yz)ln(xyz)
Which you could further simplify to (using the identity ln(a×b)=lna+lnb:
ln(xy)+ln(xz)+ln(yz)ln(xyz)=2ln(x)+2ln(y)+2ln(z)ln(xyz)
=2(ln(x)+ln(y)+ln(z))ln(xyz)
=2(ln(xyz))ln(xyz)=2
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

Hint : log(ab)=loga+logb logaa=1 logab=logbloga I am in hurry.. Please see if this can help you... If not i hope some one would help...

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