Question

Suppose a certain population increases by 20% every ten years. By what percent does the population increase each year (assuming a constant growth rate)? (Hint: the answer is not 2%.)
Show your work. Round to the nearest hundredth of a percent.

Answer 1

Let the annual growth rate as a decimal = x
\[(1+x)^{10}=1.2\]
Taking natural logs of both sides we get
\[10 \times \ln (1+x)=\ln 1.2\]
\[\ln (1+x)=\frac{\ln 1.2}{10}=0.018232\]
\[e ^{0.018232}=1+x\ .............(1)\]
Can you now solve equation (1) to find x?
When you have the value of x, multiply by 100 to convert to a decimal.

Answer 2

1.83992176?

Answer 3

Good work! When rounded to the nearest hundredth of a percent:
x = 1.84%

Answer 4

Awesome! Thanks! That actually makes sense! :)

Answer 5

You're welcome :)

Answer 6

Can you help me with this one? I'm not very good at the log things. :/

Answer 7

The expression needs to be put into the following form first;
\[\ln x ^{2}-\ln y ^{6}+\ln e ^{\frac{12}{3}}\]
Does that make sense?

Answer 8

Yes, that makes sense.

Answer 9

Good :)
Now use the following properties of logs:
\[\log_{} a+\log_{} b=\log_{} (a \times b)\]
\[\log_{} a-\log_{} b=\log_{} \frac{a}{b}\]

Answer 10

I don't think I did this right..
I got it down to log(24/18)

Answer 11

The answer must include x and y, the reason being that no values are given for these.
The first two terms can be reduced to one term as follows:
\[\ln x ^{2}-\ln y ^{6}=\ln (\frac{x ^{2}}{y ^{6}})\]
So we now have
\[\ln (\frac{x ^{2}}{y ^{6}})+\ln e ^{4}\]
Can you use the properties of logs to make this the log of one quantity?

Answer 12

ohh. woops. okay, so \[((x ^{2} \div y ^{6}) \times e ^{4}\]

Answer 13

)

Answer 14

@kropot72 ?

Answer 15

I am trying to answer, but the LATEX won't work for me now. You are on the right track.
Express it as ln (.....) and convert e^4 to 54.598.

Answer 16

ln(x^2 / y^6) 53.598
could you distribute that to be ln (53.598x^2 / 53.598y^6)?

Answer 17

Not really. Do it as
ln(54.598x^2/y^6)

Answer 18

okay. So that's the answer then?

Answer 19

Yes.

Answer 20

okay, thank you!