Question

Prove that `x^((logy-logz))y^((logz-logx))z^((logx-logy))=1`?

Answer 1

Please see the explanation.

Explanation:

Prove:

`x^(log_10(y)-log_10(z))y^(log_10(z)-log_10(x))z^(log_10(x)-log_10(y))=1`

Use the base 10 logarithm on both sides:

`log_10(x^(log_10(y)-log_10(z))y^(log_10(z)-log_10(x))z^(log_10(x)-log_10(y)))=log_10(1)`

The right side becomes 0:

`log_10(x^(log_10(y)-log_10(z)))+log_10(y^(log_10(z)-log_10(x)))+log_10(z^(log_10(x)-log_10(y)))=0`

Use the property of logarithms `log_b(a^(c-d)) = log_b(a)(c-d)`

`log_10(x)(log_10(y)-log_10(z))+log_10(y)(log_10(z)-log_10(x))+log_10(z)(log_10(x)-log_10(y))=0`

Use the distributive property on all of the parenthesis:

`log_10(x)log_10(y)-log_10(x)log_10(z)+log_10(y)log_10(z)-log_10(y)log_10(x)+log_10(z)log_10(x)-log_10(z)log_10(y))=0`

Begin canceling terms:

`cancel(log_10(x)log_10(y))-log_10(x)log_10(z)+log_10(y)log_10(z)cancel(-log_10(y)log_10(x))+log_10(z)log_10(x)-log_10(z)log_10(y))=0`

`cancel(-log_10(x)log_10(z))+log_10(y)log_10(z)cancel(+log_10(z)log_10(x))-log_10(z)log_10(y))=0`

`cancel(+log_10(y)log_10(z))cancel(-log_10(z)log_10(y))=0`

`0 = 0`

Q.E.D.

Answer 2

Please see below for the proof.

Explanation:

We will not be writing base as `10`, hence `logp=log_10p`

Now let `x^a=y^b`, then `alogx=blogy`

and `b=axxlogx/logy`

Hence `x^(logy-logz)=y^((logy-logz)xxlogx/logy)`

and `z^(logx-logy)=y^((logx-logy)xxlogz/logy)`

and hence

`x^((logy-logz))y^((logz-logx))z^((logx-logy))`

= `y^(((logy-logz)xxlogx/logy))y^((logz-logx))y^(((logx-logy)xxlogz/logy))`

= `y^[((logy-logz)xxlogx/logy)+(logz-logx)+(((logx-logy)xxlogz/logy))]`

= `y^[logx-(logzlogx)/logy+logz-logx+(logxlogz)/logy-logz]`

= `y^0=1`

Answer 3

See below.

Explanation:

`x^log(y/z) y^log(z/x) z^log(y/x)=1`

or

`log(y/z)log(x)+log(z/x)log(y)+log(y/x)log(z)=0`

or

`(y/z)^log(x)(z/x)^log(y) (x/y)^log(z)=1`

or

`x^log(z/y)y^log(x/z) z^log(y/x)=1=x^log(y/z) y^log(z/x) z^log(y/x)=x^-log(z/y)y^-log(x/z) z^-log(y/x)`

then

`x^log(z/y)y^log(x/z) z^log(y/x)=1/(x^log(z/y)y^log(x/z) z^log(y/x)`

so

`x^log(z/y)y^log(x/z) z^log(y/x)=1` is an identity.