Question 770321
Given 2^x=3^y=5^z=900^w
Show that:
xyz=2w(xy+yz+xz)
<pre>
Suppose that

2<sup>x</sup> = 3<sup>y</sup> = 5<sup>z</sup> = 900<sup>w</sup> = A

Then (you can also use log instead of ln, as above):

log(2<sup>x</sup>) = log(3<sup>y</sup>) = log(5<sup>z</sup>) = log(900<sup>w</sup>) = log(A)

x·log(2) = y·log(3) = z·log(5) = w·log(900) = log(A)

{{{x=log(A)/log(2)}}}, {{{y=log(A)/log(3)}}}, {{{z=log(A)/log(5)}}}, {{{w=log(A)/log(900)}}}

<font color="red">***</font> xyz = {{{log(A)/log(2)}}}{{{""*""}}}{{{log(A)/log(3)}}}{{{""*""}}}{{{log(A)/log(5)}}} = {{{(log(A))^3/(log(2)log(3)log(5))}}}<font color="red">***</font>

xy = {{{(log(A)/log(2))}}}{{{""*""}}}{{{(log(A)/log(3))}}} = {{{(log(A))^2/(log(2)log(3))}}}

yz = {{{(log(A)/log(3))}}}{{{""*""}}}{{{(log(A)/log(5))}}} = {{{(log(A))^2/(log(3)log(5))}}}

xz = {{{(log(A)/log(2))}}}{{{""*""}}}{{{(log(A)/log(5))}}} = {{{(log(A))^2/(log(2)log(5))}}}

xy + yz + xz = {{{(log(A))^2/(log(2)log(3))}}}{{{""+""}}}{{{(log(A))^2/(log(2)log(3))}}}{{{""+""}}}{{{(log(A))^2/(log(2)log(3))}}}{{{""=""}}}

          {{{(log(A))^2( 1/(log(2)log(3)) + 1/(log(3)log(5))+ 1/(log(2)log(5)))}}}{{{""=""}}}{{{(log(A))^2( ( log(5) + log(2)+ log(3))/(log(2)log(3)log(5)))}}}

{{{w}}}{{{""=""}}}{{{log(A)/log(900)}}}{{{""=""}}}{{{log(A)/log(2^2*3^2*5^2)}}}{{{""=""}}}{{{log(A)/(log(2^2)+log(3^2)+log(5^2))}}}{{{""=""}}}{{{log(A)/(2log(2)+2log(3)+2log(5))}}}{{{""=""}}}

          {{{log(A)/(2(log(2)+log(3)+log(5)))}}}

So 2w = {{{2log(A)/(2(log(2)+log(3)+log(5)))}}}{{{""=""}}}{{{cross(2)log(A)/(cross(2)(log(2)+log(3)+log(5)))}}}{{{""=""}}}{{{log(A)/(log(2)+log(3)+log(5))}}}

2w(xy+yz+xz){{{""=""}}}{{{log(A)/(log(2)+log(3)+log(5))}}}{{{""*""}}}{{{(log(A))^2( ( log(5) + log(2)+ log(3))/(log(2)log(3)log(5)))}}}{{{""=""}}}

          {{{log(A)/(cross(log(2)+log(3)+log(5)))}}}{{{""*""}}}{{{(log(A))^2( (cross( log(5) + log(2)+ log(3)))/(log(2)log(3)log(5)))}}}{{{""=""}}}{{{(log(A))^3/(log(2)log(3)log(5))}}}

And that's what xyz was shown equal to above, marked <font color="red">***</font>

Edwin</pre>