Every 10 years the population of a city is five-fourths of what it was 10 years before.

Question

I am working through Serge Lang's "Basic mathematics", currently on chapter 1, Section 5 question 21.

The part that troubles me is, when I asked someone I know how to solve this, they suggested using logarithm's, except in Lang's book, logarithm's are not until Chapter 13.

So obviously another way exists which I can't figure out.

What would be the best way to solve this assuming one hasn't yet learned log's?

Question:

Every 10 years the population of a city is five-fourths of what it was 10 years before.

A) How many years does it take before the population doubles?

B) Before it triples?

Answer 1

The easiest way to do this is to come up with a formula for the change in the population after $n$ years, and then try and figure out what whole number you have to plug in to $n$ so the change is atleast 2, and then atleast 3.

EDIT: the $A(t)=A_0*(1-r)^{t/p}$ is very similar to what you will have to use, but that is for some value that is decreasing by $r$ every $p$ years. So if the population of the city was decreasing by $\frac{1}{3}$ you would get a formula of $A(t)=A_0 \cdot (1-\frac {1}{3})^{t/10}=A_0 \cdot (\frac {2}{3})^{t/10}$. Can you modify that, or find the right formula, when the population is increasing?

Answer 2

Given the formula for a geometric sequence: $\{a_n\} = a_1 * r^{n}$, you can find an equation for it. Let $r = \frac54$ and call $n$ $x$ and $a_1$ is the initial population. You get: $$A = a_1*\left(\frac54\right)^{x}$$ where $x$ is the number of years passed (multiplies $a_1$ by $\frac54$ each year). To find $x$ when $A = 2a_1$: $$2a_1=a_1*\left(\frac54\right)^x$$$$2=\left(\frac54\right)^x$$ test the equation for $x = {1,2,3,4}$ (multiply $\frac54$ by itself ${1,2,3,4}$ times) and you will see that $3<x<4$ so the answer is $4$ because you are asked to find an integer amount of years. If you want to solve for the exact answer, you do have to use a logarithm:$$\log_{\frac54}2=x$$$$\frac{\log_{10}2}{\log_{10}{\frac54}} = x$$$$x\approx3.1$$