Question 779653
One way to solve this is...<br>
{{{a^(1/x)=b^(1/y)=c^(1/z)}}}
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/a = r}}}
or b = a*r
and
{{{c/b = r}}}
or c = b*r<br>
Since b = a*r, we can write c in terms of a:
{{{c = (a*r)*r = a*r^2}}}<br>
Substituting these expressions in "a" for "b" and "c" into the given equation we get:
{{{a^(1/x)=(a*r)^(1/y)=(a*r^2)^(1/z)}}}<br>
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^(1/x)=(a*r)^(1/y)}}}
First let's eliminate the fractions in the exponents. Raising both sides to the LCD power:
{{{(a^(1/x))^(x*y)=((a*r)^(1/y))^(x*y)}}}
which simplifies to:
{{{a^y=(a*r)^x}}}
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(a, (a^y))=log(a, ((a*r)^x))}}}
Using a property of logs, the exponents in the arguments can be moved out in front:
{{{y*log(a, (a))=x*log(a, (a*r))}}}
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*(log(a, (a)) + log(a, (r)))}}}
The first log on the right is a 1 so this simplifies to:
{{{y = x(1 + log(a, (r)))}}}
{{{y = x + x*log(a, (r))}}}<br>
Now we repeat the process for z:
{{{a^(1/x)=(a*r^2)^(1/z)}}}
{{{(a^(1/x))^(x*z)=((a*r^2)^(1/z))^(x*z)}}}
{{{a^z=(a*r^2)^x}}}
{{{log(a, (a^z))=log(a, ((a*r^2)^x))}}}
{{{z*log(a, (a))=x*log(a, (a*r^2))}}}
{{{z=x*(log(a, (a)) + log(a, (r^2)))}}}
{{{z=x*(1 + 2*log(a, (r)))}}}
{{{z=x + 2x*log(a, (r))}}}<br>
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[1] = y - x}}}
Substituting in for y:
{{{d[1] = (x + x*log(a, (r))) - x}}}
The x's cancel:
{{{d[1] = x*log(a, (r)))}}}
Now lets try
{{{d[2] = z - y}}}
Substituting for both z and y:
{{{d[2] = (x + 2x*log(a, (r))) - (x + x*log(a, (r)))}}}
Again the x's cancel:
{{{d[2] = 2x*log(a, (r)) - x*log(a, (r))}}}
These are like terms so we can subtract them:
{{{d[2] = x*log(a, (r))}}}<br>
As we can see, the two differences are the same. So x, y and z are in an AP.