How can I round up to the nearest power of a number?

Question

I'm sorry if I got the terminology wrong. Let me explain with an example.

Given a multiplier (we'll say 2.5), I need an equation or algorithm to find what power level a number resides in (preferably calculated, not a loop).

Level 0: < 2.5
Level 1: < 6.25 (2.5 ^ 2)
Level 2: < 15.625 (2.5 ^ 3)

And input of 8 would give an output of 15.625, as it is greater than or equal to 6.25 and less than 15.625.

Please let me know if my question is unclear. Thanks guys.

Answer 1

If $b$ is the base ("multiplier", $2.5$ in the example) and $x$ is the number to be tested, the next power of $b$ is $$b^{\left\lfloor\log_b x\right\rfloor+1}$$ where $\log_b x$ is the logarithm to base $b$ and $\lfloor x\rfloor$ is the floor function.

If you don't have access to arbitrary logarithms, use $$\log_b x = \frac{\log x}{\log b}$$ where the logarithm on the right can be whatever logarithm is available to you — typically either the natural logarithm, or the base 10 logarithm —, but it must be the same for the numerator and denominator.

Edit note: I corrected the formula after I noticed that when your number already is a power of the given base, you want the next one. The pre-edit formula would have given the same power for this case (so for example, $6.25$ would have resulted in $6.25$ with the old formula, but gives $15.625$ with the corrected one. Note that for any other numbers, both give the same result.

Answer 2

I think the formula $2.5^{\lceil \log_{2.5}x \rceil}$ is the one you want. So for your example with $x=8$ we have,

$$2.5^{\lceil \log_{2.5}x \rceil} = 2.5^{\lceil \log_{2.5}8 \rceil} = 2.5^{\lceil 2.26941\dots \rceil} = 2.5^3 = 15.625.$$

EDIT: This formula might be a bit off in certain edge cases since you have $<$'s and not $\leq$'s. In particular, if $\log_{2.5}x$ is an integer, you would want to add $1$ to exponent. Whether or not you would want to do that is moreso based on the context of the problem you are trying to solve though.