Question 1140061
ONE WAY:
We know {{{x<>0}}} , {{{y<>0}}} and {{{z<>0}}} because otherwise they would make the denominators zero, and the expression {{{1/(2x) + 1/(4y) + 1/(8z)}}} would not exist as a real number.
As {{{4=2^2}}} and {{{8=2^3}}} ,
if {{{2^x = 4^y = 8^z}}} ,
{{{2^x = 4^y = (2^2)^y = 2^(2y)}}} ,
and {{{2^(2y)/2^x=2^(2y-x)=1 meaning that {{{2y-x=0}}} , {{{x=2y}}} , and {{{y=(1/2)x}}}.
Also {{{2^x = 8^z = (2^3)^z = 2^(3z)}}} , which means {{{x=3z}}} , and {{{z=(1/3)x}}} .
You could also talk about taking logarithms base 2.\ to get to the same conclusions.
Then, substituting into {{{1/(2x) + 1/(4y) + 1/(8z) = 22/7}}} , we get
{{{1/(2x) + 1/(4(1/2)x) + 1/(8z) = 22/7}}}
{{{1/(2x) + 1/(2x) + 1/(8z) = 22/7}}}
{{{1/x + 1/(8z) = 22/7}}}
{{{1/(3z) + 1/(8z) = 22/7}}}
Multiplying both sides of the equation times the non-zero number {{{3*8*7*z}}} ,
we get the equivalent equation
{{{8*7+3*7=22*3*8*z}}}
{{{(8+3)*7=22*3*8*z}}}
{{{11*7=2*11*3*8*z}}}
{{{7/(2*3*8)=z}}}
{{{highlight(z=7/48)}}}
Then, as {{{x=3z}}} , {{{x=3*7/(2*3*8)=7/(2*8)=highlight(7/16)}}} ,
and as {{{y=(1/2)x}}} , {{{y=(1/2)*(7/16)=highlight(7/32)}}}