Exponential functions with 2 data points

Question

It's been 30 yrs since I've done this kind of math and I'm extremely weak on even the basics.

I need a function that is exponential ($y=a*b^x$) but will fit 2 data points. I think I came up with a way to develop a function but I used a calculator's "solver" function to give a decimal result rather than an expression. The problem with this is that the decimal based coefficient of the function causes y to be slightly higher than 1 if x=100,000. What would be the pure function expression for this?

• datapoint 1A) x=100,000 , y=1
• datapoint 2A) x=45,000,000 , y=450

What I did...

$y=a*b^x$

(solving for a)

1=a*b^100000

if... b=1 (because y=1 in the datapoint series and 1 is easy to work with)...

1=a*1^100,000 then a=1//

(then, solving for b)

$y=a*b^x$

450=1*b^4,500,000

(450/1)=b^4,500,000

then using solver, b=1.0000013576115

function that fits both points is: $y=1.0000013576115^x$

But if I substitute x=100,000 into this function, I get 1.1454081746 and not 1, per the original data point. I suspect it's because it would take a large number of decimal levels of precision to make y=1 when x=100,000. What would be the purely expression form of this function?

Also, I need to make a similar exponential function for these 2 data points, which is more difficult to solve because there's no y=1 to start with:

• datapoint 1B) x=100,000 , y=0.05 (5 is repeating)
• datapoint 2B) x=4,500,000 , y=25

Answer 1

The issue is in how you solved for $a$. You cannot suppose that $b=1$ just because it simplifies the formula. Part of the problem is to find the value of $b$!

So you do have two correct equations

$1 = a b^{100000}$
and
$25 = ab^{4500000}$.

Now we have two equations and two unknowns. To find $b$ you can divide the two equations to get $25=b^{4500000-100000}=b^{4400000}$ which gives
$b = 25^{1/4400000}\approx 1.00000073$. Then you can find $a$ by plugging this back into either equation. Let's take the first:

$1 = a(1.00000073)^{100000}$ which gives
$a \approx .92960085$

So $y = .92960085(1.00000073)^x$.

The same process works for your other example. The fact that one of the $y$ values was $0$ wasn't particularly helpful. However, if one of your $x$ values is $0$, the corresponding $y$ value is $a$ and so this simplifies a bit.