A lake is stocked with 2,300 young trout. After x years, the function P(x) = 2,300 e^-0.2x provides the number of the original trout that are still in the lake. When will there be 500 of the original trout left?

I got stuck... ;-;

Question

A lake is stocked with 2,300 young trout. After x years, the function P(x) = 2,300 e^-0.2x provides the number of the original trout that are still in the lake. When will there be 500 of the original trout left? 

I got stuck... ;-;

ria23 · Answer

$$P(x)=2,300e^{-0.2x}$$$$500=2300e^{-0.2x}$$
$$\frac{ 2300 }{ 500 }=4.6$$

anonymous · Answer

should be
500/2300 = e^(-0.2x)

anonymous · Answer

ln(5/23) = -0.2x
x = ln(5/23)/(-0.2)
x = 7.63 years

ria23 · Answer

Thank yhu...
How did yhu get 5/23?

anonymous · Answer

reduced 500/2300

ria23 · Answer

Oh.
Well, thank yhu for helping n_n