Can someone help me solve this please. The size of a small herbivore population at the time t(in years) obeys the function P(t)=900e^0.23t if they have enough food and the predator population stays constant. After how many years will the population reach 4500?

Question

anonymous · Answer

$$4500=900e^{0.23t}$$

mrhoola · Answer

let p(t) = 4500

mrhoola · Answer

use natural log laws to solve for t.

anonymous · Answer

so what do I do with the e?

mrhoola · Answer

first divide by 900 then use natural log to both sides of the equation 
remember ln x e = 1

anonymous · Answer

$$4500=900e^{0.23t}$$$$5=e^{0.23t}$$$$\ln5=\ln e^{0.23t}$$$$\ln5=(0.23t)\ln e$$ln e = 1$$\ln5=0.23t$$$${\ln5 \over 0.23}=t$$

anonymous · Answer

so divide 4500 by 900?

anonymous · Answer

1) divide by the coefficient (900)
2) Take the ln of each side
3) Pull the exponent down in front
4) divide both sides by everything except t. (ln e = 1)
5) Get out calculator

anonymous · Answer

Sorry I could't see you had worked out the problem it just said error. But the thing I'm still confused about is the ln, what does it mean?