Question

A lake is stocked with 1,500 young trout. If the number of the original trout alive after x years is given by the function P(x) = 1,500 e^(-0.4x), when will there be 300 of the original trout left?

Answer 1

Okay, all you really need to do is set P(x) equal to 300, since I'm sure you aren't allowed to use a calculator.
\[P(x)=1500e ^{-.4x}\]
\[300=1500e ^{-.4x}\]
Now, simply solve for x.
I assume you know the properties for Logarithms, and Natural Logarithms?...

Here are the Natural Log properties:

Answer 2

@henryrodriguez713 Not sure why calculators aren't allowed in this kind of questions.

Answer 3

\[1. \ln1=0 \]

\[2. lne=1 \]

(inverse)\[3.lne ^{x}=x\]
(one-to-one)\[4. lnx=y .... then x=y \]

Answer 4

So, divide 300 by 1500
\[\frac{ 300 }{ 1500 }\]
= \[\frac{ 1 }{ 5 }\]
\[\frac{ 1 }{ 5 }=e ^{-.4x}\]
multiply each side by ln
=\[ln\frac{ 1 }{ 5 }=lne ^{-.4x}\]
=\[(ln(\frac{ 1 }{ 5 }))={-.4x}\]
Now, divide each side by -.4
\[\frac{ (\ln (\frac{ 1 }{ 5}) }{ -.4 }=x\]
That's your answer
x=4.02359

Answer 5

@tyteen4a03 Because you could then simply analyze the graph, and not know how the answer came about. At least thats how my teacher taught me, but either way it's the same answer.

Answer 6

thank you so so so much!! really helps!!

Answer 7

Glad I could help. :)