# 1.5 Exponential and logarithmic functions  (Page 2/17)

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 $\mathbit{\text{x}}$ $1.4$ $1.41$ $1.414$ $1.4142$ $1.41421$ $1.414213$ ${\mathbf{2}}^{\mathbit{\text{x}}}$ $2.639$ $2.65737$ $2.66475$ $2.665119$ $2.665138$ $2.665143$

## Bacterial growth

Suppose a particular population of bacteria is known to double in size every $4$ hours. If a culture starts with $1000$ bacteria, the number of bacteria after $4$ hours is $n\left(4\right)=1000·2.$ The number of bacteria after $8$ hours is $n\left(8\right)=n\left(4\right)·2=1000·{2}^{2}.$ In general, the number of bacteria after $4m$ hours is $n\left(4m\right)=1000·{2}^{m}.$ Letting $t=4m,$ we see that the number of bacteria after $t$ hours is $n\left(t\right)=1000·{2}^{t\text{/}4}.$ Find the number of bacteria after $6$ hours, $10$ hours, and $24$ hours.

The number of bacteria after 6 hours is given by $n\left(6\right)=1000·{2}^{6\text{/}4}\approx 2828$ bacteria. The number of bacteria after $10$ hours is given by $n\left(10\right)=1000·{2}^{10\text{/}4}\approx 5657$ bacteria. The number of bacteria after $24$ hours is given by $n\left(24\right)=1000·{2}^{6}=64,000$ bacteria.

Given the exponential function $f\left(x\right)=100·{3}^{x\text{/}2},$ evaluate $f\left(4\right)$ and $f\left(10\right).$

$f\left(4\right)=900;f\left(10\right)=24,300.$

Go to World Population Balance for another example of exponential population growth.

## Graphing exponential functions

For any base $b>0,b\ne 1,$ the exponential function $f\left(x\right)={b}^{x}$ is defined for all real numbers $x$ and ${b}^{x}>0.$ Therefore, the domain of $f\left(x\right)={b}^{x}$ is $\left(\text{−}\infty ,\infty \right)$ and the range is $\left(0,\infty \right).$ To graph ${b}^{x},$ we note that for $b>1,{b}^{x}$ is increasing on $\left(\text{−}\infty ,\infty \right)$ and ${b}^{x}\to \infty$ as $x\to \infty ,$ whereas ${b}^{x}\to 0$ as $x\to \text{−}\infty .$ On the other hand, if $0 is decreasing on $\left(\text{−}\infty ,\infty \right)$ and ${b}^{x}\to 0$ as $x\to \infty$ whereas ${b}^{x}\to \infty$ as $x\to \text{−}\infty$ ( [link] ).

Visit this site for more exploration of the graphs of exponential functions.

Note that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.

## Rule: laws of exponents

For any constants $a>0,b>0,$ and for all x and y ,

1. ${b}^{x}·{b}^{y}={b}^{x+y}$
2. $\frac{{b}^{x}}{{b}^{y}}={b}^{x-y}$
3. ${\left({b}^{x}\right)}^{y}={b}^{xy}$
4. ${\left(ab\right)}^{x}={a}^{x}{b}^{x}$
5. $\frac{{a}^{x}}{{b}^{x}}={\left(\frac{a}{b}\right)}^{x}$

## Using the laws of exponents

Use the laws of exponents to simplify each of the following expressions.

1. $\frac{{\left(2{x}^{2\text{/}3}\right)}^{3}}{{\left(4{x}^{-1\text{/}3}\right)}^{2}}$
2. $\frac{{\left({x}^{3}{y}^{-1}\right)}^{2}}{{\left(x{y}^{2}\right)}^{-2}}$
1. We can simplify as follows:
$\frac{{\left(2{x}^{2\text{/}3}\right)}^{3}}{{\left(4{x}^{-1\text{/}3}\right)}^{2}}=\frac{{2}^{3}{\left({x}^{2\text{/}3}\right)}^{3}}{{4}^{2}{\left({x}^{-1\text{/}3}\right)}^{2}}=\frac{8{x}^{2}}{16{x}^{-2\text{/}3}}=\frac{{x}^{2}{x}^{2\text{/}3}}{2}=\frac{{x}^{8\text{/}3}}{2}.$
2. We can simplify as follows:
$\frac{{\left({x}^{3}{y}^{-1}\right)}^{2}}{{\left(x{y}^{2}\right)}^{-2}}=\frac{{\left({x}^{3}\right)}^{2}{\left({y}^{-1}\right)}^{2}}{{x}^{-2}{\left({y}^{2}\right)}^{-2}}=\frac{{x}^{6}{y}^{-2}}{{x}^{-2}{y}^{-4}}={x}^{6}{x}^{2}{y}^{-2}{y}^{4}={x}^{8}{y}^{2}.$

Use the laws of exponents to simplify $\left(6{x}^{-3}{y}^{2}\right)\text{/}\left(12{x}^{-4}{y}^{5}\right).$

$x\text{/}\left(2{y}^{3}\right)$

## The number e

A special type of exponential function appears frequently in real-world applications. To describe it, consider the following example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests $P$ dollars in a savings account with an annual interest rate $r,$ compounded annually. The amount of money after 1 year is

$A\left(1\right)=P+rP=P\left(1+r\right).$

The amount of money after $2$ years is

$A\left(2\right)=A\left(1\right)+rA\left(1\right)=P\left(1+r\right)+rP\left(1+r\right)=P{\left(1+r\right)}^{2}.$

More generally, the amount after $t$ years is

$A\left(t\right)=P{\left(1+r\right)}^{t}.$

If the money is compounded 2 times per year, the amount of money after half a year is

$A\left(\frac{1}{2}\right)=P+\left(\frac{r}{2}\right)P=P\left(1+\left(\frac{r}{2}\right)\right).$

The amount of money after $1$ year is

$A\left(1\right)=A\left(\frac{1}{2}\right)+\left(\frac{r}{2}\right)A\left(\frac{1}{2}\right)=P\left(1+\frac{r}{2}\right)+\frac{r}{2}\left(P\left(1+\frac{r}{2}\right)\right)=P{\left(1+\frac{r}{2}\right)}^{2}.$

After $t$ years, the amount of money in the account is

$A\left(t\right)=P{\left(1+\frac{r}{2}\right)}^{2t}.$

More generally, if the money is compounded $n$ times per year, the amount of money in the account after $t$ years is given by the function

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