If you have two points, for example, (2, 6) and (3, 18), how do you find the equation if you know its exponential?
I've heard about y=ab^x but I'm not sure what those variables represent.
3 Answers
You have two points on the curve to determine the two parameters $a,b$ so plug them in, giving $6=ab^2, 18=ab^3$ Can you find $a,b?$
A linear function is one which increases (or decreases) by the same amount every time the input is increased by a unit. For example, if $f$ is linear and $f(3)=5$ and $f(4)=10$, then you know that $f(5)=15$ and $f(6)=20$. More generally, the slope of this function would be 5, and you could write the equation $f(x)=mx+b$ where $m=5$ and $b=-10$.
An exponential function is similar, except that the function increases or decreases by a fixed ratio for each increment. If $g$ is exponential, and $g(3)=5$ and $g(4)=10$, then you would know that $g(5)=20$ and $g(6)=40$. The general form for an exponential function is $g(x)=ab^x$, where $a$ is the intercept (note that $g(0)=a$) and $b$ is the ratio term. In this case, we have $g(x)=\frac{5}{8}\cdot 2^x$.
In your example, the function increases by a factor of $\frac{18}{6}=3$ as the input goes from $x=2$ to $x=3$. So the ratio term is $3$, and the function has the form $g(x)=a\cdot 3^x$. You should be able to find $a$ from here.
Both these answers assume the form f(x) = ab^x; however, this presumes that the exponential function has an asymptote at x=0; it also assumes that base is important, but b can always be converted to e. Now you can also solve this problem with the more general f(x) = ae^x + c, which is still exponential, but now leaves the asymptote open to other possibilities and uses the natural base. This is useful for modeling a variety of phenomena, which are asymptotic at some max or min other than 0.
In this case, given (x1, y1) and (x2, y2):
f(x1) = y1 = ae^x1 + c
f(x2) = y2 = ae^x2 + c
To solve for a subtract c from the first equation and divide by everything left over not a. You should get:
a = (y1-c)/e^(x1 * x) (see expression for x below)
To solve for x subtract c from both equations and divide one equation by the other, and then use the natural log. Like any sloppy mathematician, out of laziness, I leave this exercise to the reader or some future enterprising person who wants to type the steps of that piece out. However, you should get:
x = ln((y2-c)/(y1-c))/(x2-x1)
