Question

The population (in millions) of a certain island can be approximated by the function:

P(x)=50 (1.05)^x

Where x is the number of years since 2000. In which year will the population reach 200 million?

Hint: An answer such as 2002.4 would represent the year 2002.

A) 2028
B)2066
C) 2002
D) 2015

Answer 1

@freckles

Answer 2

solve

200 = 50*1.05^x

x is in millions of years

Answer 3

do you know how to begin?

Answer 4

no...

Answer 5

\[200 = 50*1.05^x\]

Answer 6

divide both sides by 50 first

Answer 7

\[4 = 1.05^x\]

Answer 8

recall:
\[\ln(b^x) = x*\ln(b)\]

Answer 9

take ln of both sides

Answer 10

the whole ln thing confuses me. do i put 1.05 and ln into the calculator?

Answer 11

after dividing by 50 on both sides you are left with
\[4 = 1.05^x\]

You then take the natural log of both sides

Answer 12

\[\ln(4) = \ln(1.05^x)\]

\[\ln(4) = x*\ln(1.05)\]

Answer 13

divide both sides now by ln(1.05) to isolate x by itself

Answer 14

1.3862943611 = x(0.0487901642) Is this right?

Answer 15

\[x = \frac{ \ln(4) }{ \ln(1.05) }\]

Answer 16

I mean this: 1.3862943611 / (0.0487901642) = x Then I just divide like normal right?

Answer 17

natural log of 4, divided by natural log of 1.05, idk i dont have a calculator on me

Answer 18

x should be about 28

Answer 19

hold on...using my calculator

Answer 20

Where x is the number of years since 2000. In which year will the population reach 200 million?

The final answer is: 2000 + (the x you calculated)

Answer 21

Okay, I got roughly 2,002.8

Answer 22

i just got x = 28.413 about... so that rounds to 28

Answer 23

2000 + x = 2000 + 28

the year is 2028 when the population reaches 200 million

Answer 24

i hope you understood how logarithms were used to solve for x. Basically the property
\[\ln(b^x) = x*\ln(b)\]

Answer 25

Yeah, I got it. Thanks!

Answer 26

k, welcome