We have an answer to "Evaluate the log expression 1logxy(xyz)+1logyz(xyz)+1logzx(xyz)"

Annette Sabin

Answered question

2022-01-20

Evaluate the log expression

\frac{1}{\log_{x y} (x y z)} + \frac{1}{\log_{y z} (x y z)} + \frac{1}{\log_{z x} (x y z)}

poleglit3

Beginner2022-01-20Added 32 answers

Use the fact that

\log_{a} b = \frac{\ln b}{\ln a}

Thus

\frac{1}{\log_{x y} (x y z)} = \frac{\ln (x y)}{\ln (x y z)}

ect.

Mary Goodson

Beginner2022-01-21Added 37 answers

By using the identity

\log_{b} (a) = \frac{\ln a}{\ln b}

Your equation becomes:

\frac{\ln (x y)}{\ln (x y z)} + \frac{\ln (x z)}{\ln (x y z)} + \frac{\ln (y z)}{\ln (x y z)}

Which you could further simplify to (using the identity

\ln (a \times b) = \ln a + \ln b :

\frac{\ln (x y) + \ln (x z) + \ln (y z)}{\ln (x y z)} = \frac{2 \ln (x) + 2 \ln (y) + 2 \ln (z)}{\ln (x y z)}

= \frac{2 (\ln (x) + \ln (y) + \ln (z))}{\ln (x y z)}

= \frac{2 (\ln (x y z))}{\ln (x y z)} = 2

RizerMix

Expert2022-01-27Added 656 answers

Hint :

\log (a b) = \log a + \log b

\log_{a} a = 1

\log_{a} b = \frac{\log b}{\log a}

I am in hurry.. Please see if this can help you... If not i hope some one would help...

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