Setting up exponential equation

Question

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

Answer 1

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 $$\ f(x) = 80\left(2\right)^{\frac{4}{3}}e^{\ln\left(2^{-\frac{1}{12}}\right)x}.$$

Answer 2

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.

Answer 3

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

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