Calculating Population Growth Rate Using Log Equations

[TN17]

Homework Statement

Equation for population growth is:
P(t) = P₀(1 + R/100)^t/t₀
R --- growth rate in %
t₀--- time period
Suppose a population grew from 10 000 to 25 000 in 6 years. If time is measured in years, calculate:
a)Yearly growth rate Answer:16.5%
b) Growth rate per decade Answer360.5%

The Attempt at a Solution

For a)
25 000 = 10000(1 + R/100)⁶
2.5 = (1 + R/100)⁶
log 2.5 = 6 log [(100 + R)/100]
This is where I got stuck. I'm not sure how to isolate "R".
For b)
I wouldn't know how to solve this without knowing how to do a).

 

[eumyang]

The Attempt at a Solution

For a)
25 000 = 10000(1 + R/100)⁶
2.5 = (1 + R/100)⁶
log 2.5 = 6 log [(100 + R)/100]⁶
This is where I got stuck. I'm not sure how to isolate "R".

(You got an extra 6 on the right side.)

I wouldn't take the log of both sides. Instead, take the 6th root of both sides:
$$\sqrt[6]{2.5} = \sqrt[6]{\left( 1 + \frac{R}{100} \right)^6}$$
Can you take it from there?

 

[TN17]

(You got an extra 6 on the right side.)

I wouldn't take the log of both sides. Instead, take the 6th root of both sides:
$$\sqrt[6]{2.5} = \sqrt[6]{\left( 1 + \frac{R}{100} \right)^6}$$
Can you take it from there?

Yep :)
Thanks a lot. I never even thought of this way. I only thought of solutions in terms of log laws.

 

[TN17]

Wait... how would I do it for b)?
I did the same method as a), but put exponent 10. I didn't end up with the answer, though.
Would I still use 25 000 and 10 000? That info was only for a span of 6 years.

 

[HallsofIvy]

Wait... how would I do it for b)?
I did the same method as a), but put exponent 10. I didn't end up with the answer, though.
Would I still use 25 000 and 10 000? That info was only for a span of 6 years.

Yes, you would still use 25000 and 10000 because that's all the information you have!

I'm not sure if you copied that formula correctly- it should be
$$P(t) = P0(1 + \frac{R}{100t_0})^{t/t_0}$$
In other words, if the growth is per 10 years, you replace R/100 (which reduces R to a decimal) by R/(100*10)= R/1000 which would be the 10 year percentage reduce to "per year".

$$25000= 10000(1+ \frac{R}{1000})^{6/10}$$
$$2.5= (1+ \frac{R}{1000})^{.6}$$
Now take the ".6" root of each side. On the left you can do that with a logarithm: to solve $x^{.6}= 2.5$ take the logarithm of both sides: $.6 ln(x)= ln(2.5)$ so that $ln(x)= ln(2.5)/.6= 1.527$ so that $x= e^{1.527}= 4.6$.

 

[TN17]

Yes, you would still use 25000 and 10000 because that's all the information you have!

I'm not sure if you copied that formula correctly- it should be
$$P(t) = P0(1 + \frac{R}{100t_0})^{t/t_0}$$

Really? =\
It says
P(t) = P₀( 1 + R/100) ^t/to in the textbook, and I got the right answer with that equation for a).

 

[zgozvrm]

(You got an extra 6 on the right side.)

I wouldn't take the log of both sides. Instead, take the 6th root of both sides:
$$\sqrt[6]{2.5} = \sqrt[6]{\left( 1 + \frac{R}{100} \right)^6}$$
Can you take it from there?

... but if you wanted to take the [itex]\log[/tex] of both sides, here's how it would go:

$$\log{2.5} = 6 \log \left( 1 + \frac{R}{100} \right)$$

$$\frac{1}{6} \log{2.5} = \log \left( 1 + \frac{R}{100} \right)$$

$$\log(2.5)^{\frac{1}{6}}= \log \left( 1 + \frac{R}{100} \right)$$

$$(2.5)^{\frac{1}{6}} = \left( 1 + \frac{R}{100} \right)$$

$$\sqrt[6]{2.5} = \left( 1 + \frac{R}{100} \right)$$


which brings you back to where eumyang suggested