survey population

[rachelmaddie]

i need my work checked please.
4.
We need to see when the equations for the population of each species are equal so equate them and solve for t.
2000e^(0.05t) = 5000e^(0.02t)
(⅖)e^(0.05t) = e^(0.02t) Divide each side by 5000.
⅖ = e^(0.02t) / e^(0.05t) Divide each side by e^(0.05t)
⅖ = e^(-0.03t) Use e^a / e^b = e^(a- b)
In(⅖) / -0.03 Divide each side by -0.03
—> = 30.54

Therefore, the population of Species A will be equal to the population of Species B after 30.54 years.

 

[Deleted member 4993]

i need my work checked please.View attachment 20656
4.
We need to see when the equations for the population of each species are equal so equate them and solve for t.
2000e^(0.05t) = 5000e^(0.02t)
(⅖)e^(0.05t) = e^(0.02t) Divide each side by 5000.
⅖ = e^(0.02t) / e^(0.05t) Divide each side by e^(0.05t)
⅖ = e^(-0.03t) Use e^a / e^b = e^(a- b)
In(⅖) / -0.03 Divide each side by -0.03
—> = 30.54

Therefore, the population of Species A will be equal to the population of Species B after 30.54 years.

You found a value for 't'.

Evaluate [2000e^(0.05t)] and [5000e^(0.02t)] - using calculator. If those come out equal - your "work" is most probably correct.

 

[Otis]

i need my work checked please …
In(⅖) / -0.03 Divide each side by -0.03
—> = 30.54 …

Hi rachel. Starting with the line shown in red, you'd stopped writing equations. It would be better to not omit the solution variable, in your work. You could include the step where you took the natural logarithm of each side, also.

ln(2/5) = -0.03t \(\;\; \quad \;\) Take the natural log of each side
ln(2/5)/(-0.03) = t \(\quad\) Divide each side by -0.03
30.54 ≈ t

The species' populations will be equal after approximately 30.5 years.

If we check t=30.54 in the given exponential-growth functions, we'll find the P values are close, but not exactly the same. The difference is due to rounding. The answer is an approximation.

?

 

[Otis]

Hmm, the very end of my previous post is missing. (I ought to have caught that, in preview.)

So, if we'd like to use a calculator for verification, then we can include additional digits. Here are the outputs, using both t=30.54 and the "full" calculator result for t.

2000e^(0.05 · 30.54) = 9208.69 (rounded)
5000e^(0.02 · 30.54) = 9209.52 (rounded)

When we divide ln(2/5) by -0.03, we could use the calculator's memory feature to store the "full" result. That makes comparing the functions' output easier.

2000e^(0.05 · 30.5430243958) = 9210.0787466
5000e^(0.02 · 30.5430243958) = 9210.0787466

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