I don't understand exponential functions too well, so if you know the answer, please also show me the process I would use to come up with it.

I'm trying to figure out how to make an exponential function that doubles every five whole integers. That would mean that:

|f(1) |1     |
|f(6) |2     |
|f(11)|4     |
|f(16)|8     |

et cetera.

I know some people have suggested using a best-fit application, but I don't think those things are too accurate. Is there a tried and true process to creating a function that matches these outputs?

  • 4
    $\begingroup$ How about $2^{(x-1)/5}=\left(\sqrt[5]2\right)^{x-1}$? $\endgroup$ Mar 24, 2021 at 21:37

1 Answer 1


You want an exponential function. This means it is of the form $$ f(n)=a\cdot b^n $$ for some numbers $a,b$. You also want $f(1)=1,f(6)=2$. That's enough to solve your problem: $$ \cases{a\cdot b^1=1\\ a\cdot b^6=2} $$ Dividing equation two by equation one, we get $b^5=2$. This is exactly the equation fifth roots are made to solve: $b=\sqrt[5]2$. Inserting this into equation one, we get $a=\frac{1}{\sqrt[5]{2}}$. Thus our function must be $$ f(n)=\frac{\sqrt[5]2^n}{\sqrt[5]2} $$

  • $\begingroup$ So If I'm understanding correctly, to figure out the function, I would need to take two different outputs in ab^x form, and divide them by each other? Is that what you did there? $\endgroup$ Mar 24, 2021 at 21:50
  • $\begingroup$ @moonbeast32 Taking two different outputs in that form is what you must do. Dividing them by one another seemed like the fastest way to get at a solution, but I'm sure there are other ways. $\endgroup$
    – Arthur
    Mar 24, 2021 at 21:55
  • $\begingroup$ ok, thanks @Arthur that helps a lot. $\endgroup$ Mar 24, 2021 at 22:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .