If p is a whole number, for what values of p is 10 x 2^p-1 < 165 with working pls?

10 × 2^(p - 1) < 165

As 10 > 0:
10 × 2^(p - 1) / 10 < 165 / 10
2^(p - 1) < 16.5
log[2^(p - 1)] < log(16.5)
(p - 1) log(2) < log(16.5)

As log(2) > 0:
(p - 1) log(2) / log(2) < log(16.5) / log(2)
p - 1 < log(16.5) / log(2)
p - 1+ 1 < [log(16.5) / log(2)] + 1
p < [log(16.5) / log(2)] + [log(2) / log(2)]
p < [log(16.5) + log(2)] / log(2)]
p < log(16.5 × 2) / log(2)
p < log(33) / log(2)

OR:
p < log₂(33)

We observe: 10x2^(p-1) is increasing function of p. That is If p is growing also 2^(p-1) is growing.
In this case it is sufficient to find the first whole number q that makes the inequality false to make sure that all the major numbers of q * make it false.
2^(p-1)< 16.5 divided both sides by 10.
We know
16=2⁴≤16.5≤2⁵=32
So q=5
In other words inequality is true for exponent ≤4.
p-1≤4
p≤5

The list of the whole number is 0, 1, 2, 3, 4, 5.

10*2^(p-1) < 165
2^(p-1) < 16.5
2^p < 33
Recall 2^5 = 32
p <= 5