help

[high mulch]

Anyone?

[errant oracle]

As we have exponential growth from the year t0 = 2007, we can write the amount at year n as:
N(t) = N0 a^(t - t0), where N0 = N(t0) and a is a constant we don't know.
Then, suppose we have the amount for two values of t:
N1 = N(t1) = N0 a^(t1 - t0)
N2 = N(t2) = N0 a^(t2 - t0)
Let's take the logarithm of each equation.
ln(N1) = ln(N0) + (t1 - t0) ln(a)
ln(N2) = ln(N0) + (t2 - t0) ln(a)
Now, let's multiply the first equation by (t2 - t0) and the second by (t1 - t0).
(t2 - t0)ln(N1) = (t2 - t0)ln(N0) + (t2 - t0)(t1 - t0) ln(a)
(t1 - t0)ln(N2) = (t1 - t0)ln(N0) + (t2 - t0)(t1 - t0) ln(a)
Now, let's subtract the second equation from the first.
(t2 - t0)ln(N1) - (t1 - t0)ln(N2) = (t2 - t1)ln(N0)
From here we get:
ln(N0) = ((t2 - t0)ln(N1) - (t1 - t0)ln(N2))/(t2 - t1)
So, now you can get N0 from here by potentiating and substituting the given values.

[high mulch]

Wha

[errant oracle]

What?

[high mulch]

I don’t get it

[errant oracle]

Which part?

We are just performing algebraic manipulation to solve the following system for N0 and a:
N1 = N0 a^(t1 - t0)
N2 = N0 a^(t2 - t0)
But since we don't need a specifically, we can just find N0 and stop there.

[high mulch]

What’s the answer then?

[errant oracle]

As I said, potentiate the last equation and substitute the given values.

[high mulch]

I got it wrong

[errant oracle]

What did you get?

[high mulch]

Nvm I got it