the letter e and exponetail decay

[shellez100]

hello, i am having trouble with natural logs in written problems, and decay problems please help where am i going wrong am i taking the right steps?


e formula models the population of a particular city
in thousands, t years after 1990. When will the population of the city reach 900 thousand?
A=340e^o.o36t
t= 900
900*.036*ln
34.2*340
11628
this is completly not the answer can anyone help?

 

[soroban]

Hello, shellez100!

You're misreading the problem.

The formula models the population of a city in thousands, \(\displaystyle t\) years after 1990.
When will the population of the city reach 900 thousand?

. . \(\displaystyle A \:=\:340e^{0.036t}\)

We are given: \(\displaystyle A = 900,\) and we want to find \(\displaystyle t.\)


\(\displaystyle 900 \:=\:340e^{0.036t} \quad\Rightarrow\quad e^{0.036t} \:=\:\frac{900}{34)} \quad\Rightarrow\quad e^{0.036t}\:=\:\frac{45}{17}\)

\(\displaystyle \text{Take logs: }\:\ln\left(e^{0.036t}\right) \:=\:\ln\left(\frac{45}{17}\right) \quad\Rightarrow\quad 0.036t\underbrace{\ln(e)}_{\text{This is 1}} \:=\:\ln\left(\frac{45}{17}\right) \)

. . \(\displaystyle 0.036t \:=\:\ln\left(\frac{45}{17}\right) \quad\Rightarrow\quad t \:=\:\dfrac{\ln(\frac{45}{17})}{0.036} \:=\:27.04025405\)


It will take about 27 years. .The year will be 2017.