If x to the power 1/p=y to the power 1/q=z to the power 1/r and xyx=1 ...
Solution:
Given:
x^(1/p) = y^(1/q) = z^(1/r)
xyz = 1
Let us assume:
k = x^(1/p) = y^(1/q) = z^(1/r)
Then k^p = x, k^q = y, and k^r = z
Substituting these values in the equation xyz=1, we get:
k^(p+q+r) = 1
Taking the logarithm of both sides, we get:
(p+q+r) log(k) = 0
Since k is positive, log(k) is also positive. Therefore, (p+q+r) must be equal to zero.
Hence, p+q+r=0.
We also know that k = x^(1/p) = y^(1/q) = z^(1/r). Therefore, we can write:
k^(p+q+r) = x^(q+r) y^(p+r) z^(p+q) = 1
Substituting p+q+r=0, we get:
x^(q+r) y^(p+r) z^(p+q) = x^(-p) y^(-q) z^(-r)
Multiplying both sides by x^p y^q z^r, we get:
x^q y^r z^p = 1
This implies that x, y, and z are the roots of the equation:
t^3 = 1
Therefore, x = y = z = 1.
Substituting this in p+q+r=0, we get:
p+q+r=0
Since x=y=z=1, we have:
k = x^(1/p) = y^(1/q) = z^(1/r) = 1
Therefore, p=q=r=0.
Hence, the values of p,q, and r are 0.
To make sure you are not studying endlessly, EduRev has designed CA Foundation study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in CA Foundation.