# How to solve for a number if its exponent is less than 1 but greater than 0 like 1/2

First off, I know how to solve for negative exponents. The inquiry is how to solve for the number if the exponent is like 0.5 or 1/2. I found this forum post which talked about the reasons it is possible and that it works.

How does an exponent work when it's less than one?

It is nice to understand that it works however I am having a hard time trying to figure out the exact math you need to do; to solve for like, $$\\36^.5$$ I know I can just use a calculator however I am trying to actually make a calculator and I need to know how to do this to incorporate this into it.

• When you say you're trying to make a calculator, I assume you're writing a computer program. Virtually any computer language has a math library that will do this for you. Some have exponentiation operators. In python, you can jut write x**(1/2) for $x^{1/2}$ – saulspatz Dec 3 '19 at 23:16
• Yes, I am making a computer program. I know I can use a program library however I would love to be able to know the mathematics to do it so I can practice other things. Thanks for the tip though! – Annonymous Dec 3 '19 at 23:57
• Have you taken calculus? – saulspatz Dec 4 '19 at 0:09
• I have not taken calculus. Is there something in it that may help by chance? If so that would be great to know. – Annonymous Dec 4 '19 at 0:19
• Even the meaning of $a^b$ can't really be made precise without calculus. Also, the numerical method of calculating the value for specific $a$ and $b$ depends on calculus, as well as numerical analysis, and an understanding of computer hardware. Even once you learn enough to be able to do this for yourself, don't do it! There are many pitfalls. The math libraries were not only written by experts, they have been debugged by millions of users over many years. I don't want to discourage you from learning this stuff. It will stand you in good stead, but use it for better purposes. – saulspatz Dec 4 '19 at 0:25

If the exponent that you want to solve for is a rational number (meaning it can be written as a fraction) it is solvable/approximate-able.

To do this, you need to understand the properties of exponential functions. One such property says that $$a^{bc} = (a^b)^c$$. When the exponent is rational and can be written as a fraction, for example, $$\frac23$$, it can be rewritten as $$2\frac13$$ and split into 2 math operations.

First, you take the number a that you want to put to the power of$$\frac23$$ and put it to the power of the first number b or in this case $$2$$ and do $$a^2$$, and then take that number and raise it to the power of $$\frac13$$.

As it says in the original forum, taking something to a fraction power $$\frac1c$$ is the same as taking the $$c^{th}$$ root of that number so the final product of raising $$a^{(\frac23)}$$ is the same as the $$3^{rd}$$ root of $$a^2$$.

And you cant really take the root of a number exactly because most of the time it will be irrational (not expressed as a fraction), but you can approximate it using calculus, or to figure out what numbers it is between numerically.