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

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.

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#

#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.