SOLUTION: x^24=w y^40=w xyz^12=w xyz=/=1 x,y,z >1 x= y= w= z=

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Question 1201476: x^24=w
y^40=w
xyz^12=w
xyz=/=1
x,y,z >1
x=
y=
w=
z=
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


x%5E24=w --> x=w%5E%281%2F24%29

y%5E40=w --> y=w%5E%281%2F40%29

Then

xyz%5E12=w
%28w%5E%281%2F24%29%29%28w%5E%281%2F40%29%29%28z%5E12%29=w
%28w%5E%285%2F120%29%29%28w%5E%283%2F120%29%29%28z%5E12%29=w
%28w%5E%288%2F120%29%29%28z%5E12%29=w
%28w%5E%281%2F15%29%29%28z%5E12%29=w
z%5E12=w%5E%2814%2F15%29
z=w%5E%2814%2F180%29=w%5E%287%2F90%29

The system of equations is indeterminate; there is an infinite family of solutions.

ANSWERS:

w=p%5E90%29 where p is any number (except 1, according to the problem description)

x=w%5E%281%2F24%29=p%5E%2890%2F24%29=p%5E%2815%2F4%29
y=w%5E%281%2F40%29=p%5E%2890%2F40%29=p%5E%289%2F4%29
z=w%5E%287%2F90%29
z%5E12=w%5E%2884%2F90%29=p%5E84

CHECK:
xyz%5E12=%28%28p%5E%2815%2F4%29%29%28p%5E%289%2F4%29%29%28p%5E84%29%29=p%5E90=w

x^24=w --> x=w^(1/24)

y^40=w --> y=w^(1/40)

Then

xyz^12=w
(w^(1/24))*(w^(1/40))*(z^12)=w
(w^(5/120))*(w^(3/120))*(z^12)=w
(w^(8/120))*(z^12)=w
(w^(1/15))*(z^12)=w
z^12=w^(14/15)
z=w^(14/180)=w^(7/90)

The system of equations is indeterminate; there is an infinite family of solutions.

ANSWERS:

w=p^90 where p is any number (except 1, according to the problem description)

x=w^(1/24)=p^(90/24)=p^(15/4)
y=w^(1/40)=p^(90/40)=p^(9/4)
z=w^(7/90)
z^12=w^(84/90)=p^84

CHECK:
xyz^12=((p^(15/4))*(p^(9/4))*(p^84))=(p^6)(p^84)=p^90=w