Maths - bouncing ball question

I can't see where I'm going wrong here:

A rubber ball is dropped from a height of 6m. It falls vertically before hitting the ground and bouncing straight back up. After the first time it hits the ground, it ascends to a height of 5.52m. The maximan height that the ball reaches after each bounce form a geometric sequence.

Show algebraically that the first time the ball bounces to a maximan height of less than one metre is after the 22nd bounce. The first time I attempted this I did this:
Un = a(r^n)
a=6
r=0.92
n=number of bounces

1> 6 x 0.92^n
1/6 > 0.92^n

then I took log base 0.92 of both sides but this gave
21.49>n

I managed to get the the answer another way but I just wondered why my original method wasn't working?

I can't see where I'm going wrong here:

A rubber ball is dropped from a height of 6m. It falls vertically before hitting the ground and bouncing straight back up. After the first time it hits the ground, it ascends to a height of 5.52m. The maximan height that the ball reaches after each bounce form a geometric sequence.

Show algebraically that the first time the ball bounces to a maximan height of less than one metre is after the 22nd bounce. The first time I attempted this I did this:
Un = a(r^n)
a=6
r=0.92
n=number of bounces

1> 6 x 0.92^n
1/6 > 0.92^n

then I took log base 0.92 of both sides but this gave
21.49>n

I managed to get the the answer another way but I just wondered why my original method wasn't working?

ik this is late but its cause its 6(0.92)^n-1 not 6(0.92)^n so the logs would be n log (0.92) - log (0.92) = log (1/6) and n would be estimated around 22.48.

The height reached after the nth bounce is not 6(0.92)^(n-1); it should be obvious this formula gives the wrong result when N=1.

As mqb said, the inequality should be reversed so you have n > 21.49; then the smallest integer value for n satisfying this is n=22.