Steve S. answered • 03/17/14
Tutor
5
(3)
Tutoring in Precalculus, Trig, and Differential Calculus
"Find four geometric means between 243 and 1."
The geometric mean of two numbers x and y, where x>0 and y>0, is √(xy).
1 < √(xy) < 243
1 < xy < (3^5)^2
1/x < y < 3^10/x
So any point between the graphs of y = 1/x, x > 0, and y = 3^10/x, x > 0, will be a valid pair of numbers that we can use to find a geometric mean to satisfy the problem.
See http://www.wyzant.com/resources/files/265437/geometric_means.
The geometric mean of two numbers x and y, where x>0 and y>0, is √(xy).
1 < √(xy) < 243
1 < xy < (3^5)^2
1/x < y < 3^10/x
So any point between the graphs of y = 1/x, x > 0, and y = 3^10/x, x > 0, will be a valid pair of numbers that we can use to find a geometric mean to satisfy the problem.
See http://www.wyzant.com/resources/files/265437/geometric_means.
