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Exponential growth versus linear growth II

Alignments to Content Standards: F-LE.A.3


Using a scientific calculator, Alex makes the following table listing values of $(1.001)^x$ and $2x$ for a few inputs:

$x$ $(1.001)^x$ $2x$
1 1.001 2
10 1.01004512 20
50 1.05124483 100
100 1.10511570 200
500 1.64830942 1000

Alex concludes from the table that the values of $2x$ grow faster than the values of $1.001^{x}$ so that $$ 2x > (1.001)^x $$ for all positive values of $x$. Is Alex correct? Explain how you know.

IM Commentary

Exponential functions with a base larger than one eventually exceed in value any given linear (or polynomial) function if the value of the input is sufficiently large. When the base is very close to 1, however, the exponential function will look like a linear function for small values of $x$: $(1+a)^x$ is approximately equal to $1 + a x$ for small values of $x$. The closer $a$ is to zero, the larger the range of values of $x$ for which this approximation applies. In this problem, $a = \frac{1}{1000}$ and once $x$ is 500, the value of $(1.001)^x$ is noticeably different from $1 + \frac{x}{1000}$.

Although a calculator can and should be used on this problem, more value will be gained from the task by thinking about the function values than by simply having a calculator plot graphs. This aspect of the task is emphasized in the second solution.


Solution: 1. Continuing the table

Notice in the table that for the smaller values of $x$ which Alex has plugged in, $1.001^x$ grows roughly like $$ 1 + \frac{x}{1000}. $$ If this behavior were to persist, then $2x$ would always remain larger than $1.001^x$ because its slope, two, is much steeper. However, toward the end of the table, it appears as if the values of $1.001^x$ are beginning to grow a little faster. Plugging in some more values of $x$ confirms this trend and shows that in fact $1.001^x$ eventually takes a larger value than $2x$, once $x$ is large enough.

$x$ $1.001^x$ $2x$
1000 2.71692393 2000
5000 148.0428362 10000
10000 21916.68134 20000

Continuing the table would reveal that the accelerating growth of $1.001^x$ persists as larger and larger $x$ values are plugged in.

Solution: 2. Estimation

A scientific calculator is not necessary to find a value of $x$ where $$ 1.001^x > 2x. $$ Write $1.001$ as $1 + \frac{1}{1000}$ and observe that $$ \left(1 + \frac{1}{1000}\right)^{1000} > 2. $$ To see this, note that the expanded form of $\left(1+\frac{1}{1000}\right)^{1000}$ contains a one (coming from the product of one thousand ones) and also one thousand terms of the form $1^{999} \times \frac{1}{1000} = \frac{1}{1000}$, coming by choosing $\frac{1}{1000}$ from one of the thousand factors and $1$ from the other 999 factors. All of the other terms are positive and so we have shown the inequality. Raising both sides of the inequality to the $n^{th}$ power gives $$ \left(1 + \frac{1}{1000}\right)^{1000n} > 2^n. $$

Choose $x = 1000n$ where $n$ satisfies $$ 2^n > 2000n. $$ Then we have \begin{eqnarray*} \left(1 + \frac{1}{1000}\right)^{x} &=& \left(\left(1 + \frac{1}{1000}\right)^{1000}\right)^n\\ &\geq& 2^n \\ &>& 2000n\\ &=& 2x. \end{eqnarray*} Recall that $2^{10}$ is a bit over 1000 (1024), and that $2^5=32$, so $2^{15}$ clearly exceeds $1000 \cdot 2 \cdot 15 = 2000 \cdot 15 =30,000$. So, $2^n> 2000n$ when $n = 15$. It then follows, choosing $x = 1000n$, that $$ \left(1+ \frac{1}{1000}\right)^{15000} > 2 \cdot 15000. $$ This value of $x$ is larger than the one found with solution 1 but did not require a calculator to find.