Question

A country's population in 1990 was 134 million. In 2000, 139 million. Estimate the population in 2014 with P=Ae^kt, while rounding k to the nearest 4 decimals.

Answer 1

Think of the year 1990 as time \(t=0\). Then according to the model, \(P=Ae^{kt}\), when \(t=0\) the population is \(P=A=134\) (ignore the "million" for now; you can always just multiply by it later).

2000 is ten years after 1990, so \(t=10\) represents the year 2000. You want to find the relative growth factor \(k\) by solving this equation:
\[\large139=134e^{10k}\]

Answer 2

That's where I keep getting stuck. Going from thee, I'm getting 14e^(3.6621*24) which is giving me something like 1.9834, but shouldn't it be giving me an answer around 146-7?

Answer 3

i mean 134 not 14

Answer 4

\[\large\begin{align*}
139&=134e^{10k}\\\\
\frac{139}{134}&=e^{10k}\\\\
\ln\frac{139}{134}&=10k\\\\
k&=\frac{1}{10}\ln\frac{139}{134}\\\\
k&\approx\cdots\end{align*}\]

Answer 5

K would be about 1.5124?

Answer 6

No, it's a lot smaller than that... Make sure you're typing it properly into your calculator.

Try (1/10)*ln(139/134)

Answer 7

I already skipped the question, but there's another one like it with 156 million in 1990, 162 mil in 1996, est it in 2016. I'm at (1/6)*In(1.038), but I keep getting 6.216 which seems way too big of a number

Answer 8

At \(t=0\), \(P=A=156\) (million, but again you can ignore that detail). At \(t=6\), \(P=162\).

So you have
\[\large 162=156e^{6k}~~\Rightarrow~~k=\frac{1}{6}\ln\frac{162}{156}\approx\cdots\]
Once you find \(k\), plug it into this eq. and solve for \(P\):
\[\large P=156e^{16k}\]
Do you have the right value of \(k\)? I'm getting about 0.00629, and a final population of about 173.

Answer 9

I think I was reading the calculator wrong or something, wasn't taking in the E-03 at the end of the number. I think I might be able to get it from here

Answer 10

Yeah the E-03 is notation for \(\#\times10^{-3}\). So if you have \(6.216\text{E}(-3)\), that's equivalent to \(6.21\times10^{-3}=.00621\).