The population of algebra was 34.9 million in 2009 and has been growing by about 2.6% each year....

Question:

The population of algebra was 34.9 million in 2009 and has been growing by about 2.6% each year. If this trend continues, when will the population exceed 45 million?

Exponential Growth:

Suppose that {eq}Q(t) {/eq} is some quantity that varies over time. We say that {eq}Q(t) {/eq} grows exponentially if there is some constant {eq}k > 1 {/eq} such that {eq}Q(t+1)=kQ(t) {/eq} for any {eq}t {/eq}. A quantity {eq}Q(t) {/eq} that grows exponentially in this way satisfies the equation {eq}Q(t)=Q(0)k^t {/eq}.

Answer and Explanation:

Let {eq}P(t) {/eq} be the population {eq}t {/eq} years after 2009, in millions of people. If the population is growing by 2.6% each year, then {eq}P(t+1)=(1+0.026)P(t)=1.026P(t) {/eq} for any {eq}t {/eq}. That is, the population is growing exponentially, and {eq}P(t)=P(0)(1.026^t) {/eq}. Since the population in 2009 was 34.9 million, {eq}P(0)=34.9 {/eq} and so {eq}P(t)=34.9(1.026^t) {/eq}.

We want to find when the population reaches 45 million: that is, when {eq}P(t)=45 {/eq}. This equation can be solved for {eq}t {/eq} as follows:

{eq}\begin{align*} P(t)&=45\\ 34.9(1.026^t)&=45\\ 1.026^t&=\frac{45}{34.9}\\ t \ln 1.026&=\ln \frac{45}{34.9}&&\text{(taking the logarithm of both sides)}\\ t&=\frac{\ln(45/34.9)}{\ln 1.026}\\ &\approx 9.9025 \, . \end{align*} {/eq}

So the population will exceed 45 million after nearly 10 years: that is, roughly in the year {eq}\boxed{2019}\, {/eq}.


Learn more about this topic:

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Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
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