# Setting up exponential equation

I want to devise a function $$f(x)$$ such that $$f(4) = 160, f(16)=80, f(28)=40$$ and $$f(40)=20$$, as well as all the other values in between $$4-40$$. As in, the numbers between $$4-12$$ have a proportional value between $$160-80$$. Numbers between $$28-40$$ have a proportional value between $$40-20$$ etc.

The values on the left go up linearly and the values on the right go down exponentially.

I hope this makes sense. But would appreciate any help. Thanks

• Are you sure the numbers on the left are correct? Because you say they go up linearly, but as you've written them they do not. Dec 7, 2020 at 19:30
• Sorry, linearly was maybe the wrong term. I was just picturing a graph with the numbers right logarithmically proportional to the numbers on the left Dec 7, 2020 at 20:15
• But the numbers on the left do not go up linearly. $4 +\color{blue}8 = 12; 12 + \color{red}{16}=28; 28+\color{green}{12}= 40$. $8, 16, 12$ are different numbers. Dec 7, 2020 at 20:16
• ah sorry. That was daft. It's meant to be 16 not 12. I just changed it there. Yeah so they do go up linearly each time by a step of 12 Dec 7, 2020 at 20:18
• Ah.... now we are talking! Dec 7, 2020 at 20:20

Since your $$x$$-values are spaced linearly and your $$y$$-values are spaced exponentially, we can find an exponential function based solely on the first two points and it will hold for all of them. Let $$f(x) = a(b)^x$$ for unknown $$a,b$$. Based on your first two points, we have \begin{align} 160 &= a(b)^{4}\\ 80 &= a(b)^{16}. \end{align} Dividing the first equation by the second lets us eliminate $$a$$: $$2 = b^{-12} \implies b = 2^{-1/12}.$$

Plugging this back in to either equation (I'll use the second): $$80 = a(2^{-1/12})^16) = a(2)^{-16/12} = a(2^{-4/3}) \implies a = 80(2)^{4/3}.$$

Putting it all together gives us $$f(x) = 80(2)^{4/3}(2^{-1/12})^{x},$$ which if we want we can rewrite as $$f(x) = 80(2)^{\frac{4}{3} - \frac{x}{12}} = 80(2)^{\frac{16-x}{12}}.$$

If for some reason you wanted to use base $$e$$ instead this would become $$\ f(x) = 80\left(2\right)^{\frac{4}{3}}e^{\ln\left(2^{-\frac{1}{12}}\right)x}.$$

• Damn that's good. I wish I could do that. I mean I get it! But I couldn't have worked it out myself. Thanks a lot. I'll put this to good use! Dec 7, 2020 at 20:32
• It's not all that bad, just takes some practice. Dec 7, 2020 at 20:33

So $$Left = 4 + 12(t-1)$$ or $$-8 + 12t$$. And $$Right = 160\cdot (\frac 12)^{t-1}$$ or $$320\cdot (0.5)^t$$.

It's probably easiest to leave it as that $$x(t) = Left = 12t -8$$ while $$y(t) = Right = 360\cdot (0.5)^{t}$$.

But if you must have $$Right(Left) =$$some function in terms of $$Left$$ then:

Solve $$t$$ in terms of $$Left$$ and plug it into the formula for $$Right$$.

$$Left = -8 + 12t$$

$$12t = Left +8$$

$$t = \frac {Left + 8}{12}$$

$$Right = 360\cdot (0.5)^{t}= 360\cdot (0.5)^{\frac {Left+8}{12}}$$

If you like you can shift the offset:

$$0.5^{\frac {Left+8}{12} } = 0.5^{\frac {Left}{12}+\frac 23}=[\frac 12]^{\frac 23}0.5^{\frac {Left}{12}}$$ s

$$Right = [\frac {360}{2^{\frac 23}}]\cdot (0.5)^{\frac {Left}{12}}$$

• Ah interesting! Probably a little trickier to grasp than the other solution but think I think I get it. Really appreciate the response! Dec 7, 2020 at 20:36

Take the logarithm of the values on the right column and plot them vs. the values on the left column.

You should obtain quite a straight line.
in any case draw the line that best approximate the graph and you have the parameters of the exponential function you are looking for.

• Okay, I need it in the form of a function for a max/msp patch i'm creating. As in a algebraic function so I can put multiple values through it Dec 7, 2020 at 20:14