A ball is dropped from a height of 9 feet and allowed to bounce. Each time the ball bounces, it reaches a height that is23the height of the previous bounce. How many times will the ball be exactly 3 feet above the ground? ​

I'm assuming you meant to say "it reaches a height that is 2/3 the height of the previous bounce".

We have a geometric sequence with a = 9 as the first term and r = 2/3 as the common ratio.

The nth term is a*(r)^(n-1) = 9*(2/3)^(n-1)

Set this equal to 3 and solve for n. You'll need logs to isolate the exponent.

9*(2/3)^(n-1) = 3

(2/3)^(n-1) = 3/9

(2/3)^(n-1) = 1/3

log[ (2/3)^(n-1) ] = log(1/3)

(n-1)*log(2/3) = log(1/3)

(n-1)*(-0.17609) = -0.47712

n-1 = (-0.47712)/(-0.17609)

n-1 = 2.7095235

n = 2.7095235+1

n = 3.7095235

Unfortunately we don't get a whole number. We can see that the terms of the sequence are:

9, 6, 4, 2.667

Each term is found by multiplying 2/3 by the previous term.

We don't land exactly on "3". The closest we get is 2.667 approximately.

So to answer the question, there is no way to get to exactly 3 feet above the ground based on a whole number of bounces.