SOLUTION: a^(1/x)=b^(1/y)=c^(1/z) if a,b,c are in GP. Prove x,y,z are in AP

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Question 779653: a^(1/x)=b^(1/y)=c^(1/z) if a,b,c are in GP. Prove x,y,z are in AP
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
One way to solve this is...

a%5E%281%2Fx%29=b%5E%281%2Fy%29=c%5E%281%2Fz%29
If a, b and c are in a geometric progression (which I assume "GP" means) then consecutive terms have a common (fixed) ratio. If we call this ratio "r" then
b%2Fa+=+r
or b = a*r
and
c%2Fb+=+r
or c = b*r

Since b = a*r, we can write c in terms of a:
c+=+%28a%2Ar%29%2Ar+=+a%2Ar%5E2

Substituting these expressions in "a" for "b" and "c" into the given equation we get:
a%5E%281%2Fx%29=%28a%2Ar%29%5E%281%2Fy%29=%28a%2Ar%5E2%29%5E%281%2Fz%29

To see if x, y and z are in an arithmetic progression (AP), where consecutive terms have a common (fixed) difference, we will start by expressing y and z in terms of x. First we'll do y:
a%5E%281%2Fx%29=%28a%2Ar%29%5E%281%2Fy%29
First let's eliminate the fractions in the exponents. Raising both sides to the LCD power:
%28a%5E%281%2Fx%29%29%5E%28x%2Ay%29=%28%28a%2Ar%29%5E%281%2Fy%29%29%5E%28x%2Ay%29
which simplifies to:
a%5Ey=%28a%2Ar%29%5Ex
Now we use logarithms to get the x's and y's out of the exponents. Finding the base a log of each side:
log%28a%2C+%28a%5Ey%29%29=log%28a%2C+%28%28a%2Ar%29%5Ex%29%29
Using a property of logs, the exponents in the arguments can be moved out in front:
y%2Alog%28a%2C+%28a%29%29=x%2Alog%28a%2C+%28a%2Ar%29%29
The log on the left is just a 1. On the right we can use another property of logs to split it into two logs (separating the "a" and the "r"):
y+=+x%2A%28log%28a%2C+%28a%29%29+%2B+log%28a%2C+%28r%29%29%29
The first log on the right is a 1 so this simplifies to:
y+=+x%281+%2B+log%28a%2C+%28r%29%29%29
y+=+x+%2B+x%2Alog%28a%2C+%28r%29%29

Now we repeat the process for z:
a%5E%281%2Fx%29=%28a%2Ar%5E2%29%5E%281%2Fz%29
%28a%5E%281%2Fx%29%29%5E%28x%2Az%29=%28%28a%2Ar%5E2%29%5E%281%2Fz%29%29%5E%28x%2Az%29
a%5Ez=%28a%2Ar%5E2%29%5Ex
log%28a%2C+%28a%5Ez%29%29=log%28a%2C+%28%28a%2Ar%5E2%29%5Ex%29%29
z%2Alog%28a%2C+%28a%29%29=x%2Alog%28a%2C+%28a%2Ar%5E2%29%29
z=x%2A%28log%28a%2C+%28a%29%29+%2B+log%28a%2C+%28r%5E2%29%29%29
z=x%2A%281+%2B+2%2Alog%28a%2C+%28r%29%29%29
z=x+%2B+2x%2Alog%28a%2C+%28r%29%29

If x, y and z are in an AP then they should have a common difference (which we will call "d"). Let's see:
d%5B1%5D+=+y+-+x
Substituting in for y:
d%5B1%5D+=+%28x+%2B+x%2Alog%28a%2C+%28r%29%29%29+-+x
The x's cancel:
d%5B1%5D+=+x%2Alog%28a%2C+%28r%29%29%29
Now lets try
d%5B2%5D+=+z+-+y
Substituting for both z and y:
d%5B2%5D+=+%28x+%2B+2x%2Alog%28a%2C+%28r%29%29%29+-+%28x+%2B+x%2Alog%28a%2C+%28r%29%29%29
Again the x's cancel:
d%5B2%5D+=+2x%2Alog%28a%2C+%28r%29%29+-+x%2Alog%28a%2C+%28r%29%29
These are like terms so we can subtract them:
d%5B2%5D+=+x%2Alog%28a%2C+%28r%29%29

As we can see, the two differences are the same. So x, y and z are in an AP.