Solve a in a power n equal b

Class room blackboard with mathematics equations written in chalk.

Finding the value of a in equation (1):

an = b (1)

can be resolved using logarithms.


First, transform equation (1) to equation (2):

ln(an) = ln(b) (2)


Now we can easily extract the power (n) by changing it into a multiplication:

n × ln(a) = ln(b) (3)


From equation (3), we know what the logarithm of a is equal to:

ln(a) = ln(b) over n (4)


Finally, we can write an equation as a = something.

a equal e power ln(b) over n (5)


Now we can calculate a in any equation such as the following which could be used to know how much percent you need to multiply your money by 2 in 12 months:

a12 = 2 (6)

The answer is a ≈ 1.059463094. If you can find a system where you make around 5.95% per month, cumulatively, then at the end of the year, you will have multiplied your money by 2 since:

  1.059512 = 2.000836183

Note that equation (3) is also useful to determine n if you have a and b:

n equal ln(b) over ln(a) (7)

Equation (5) is why you see so many equations with e power something. We're calculating an a of some sort in those equations. This is also why I used the ln() function, which is called the Neperian logarithms. The letter e is the based used by the Neperian logarithms. The log() function uses base 10 by default. You can also specify the base by adding a subscript as in equation (8) which is used to determine the number of digits in a binary number:

floor(log(base 2) of (n)) = floor(ln(n) over ln(2)) (8)

Notice how this is equivalent to using any base logarithm (Neperian in my example) and dividing by that same base logarithm of the base.