# 1.5 Exponential and logarithmic functions  (Page 3/17)

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

What happens as $n\to \infty ?$ To answer this question, we let $m=n\text{/}r$ and write

${\left(1+\frac{r}{n}\right)}^{nt}={\left(1+\frac{1}{m}\right)}^{mrt},$

and examine the behavior of ${\left(1+1\text{/}m\right)}^{m}$ as $m\to \infty ,$ using a table of values ( [link] ).

 $\mathbit{\text{m}}$ $10$ $100$ $1000$ $10,000$ $100,000$ $1,000,000$ ${\mathbf{\left(}\mathbf{1}\mathbf{+}\frac{\mathbf{1}}{\mathbit{\text{m}}}\mathbf{\right)}}^{\mathbit{\text{m}}}$ $2.5937$ $2.7048$ $2.71692$ $2.71815$ $2.718268$ $2.718280$

Looking at this table, it appears that ${\left(1+1\text{/}m\right)}^{m}$ is approaching a number between $2.7$ and $2.8$ as $m\to \infty .$ In fact, ${\left(1+1\text{/}m\right)}^{m}$ does approach some number as $m\to \infty .$ We call this number $e$    . To six decimal places of accuracy,

$e\approx 2.718282.$

The letter $e$ was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although Euler did not discover the number, he showed many important connections between $e$ and logarithmic functions. We still use the notation $e$ today to honor Euler’s work because it appears in many areas of mathematics and because we can use it in many practical applications.

Returning to our savings account example, we can conclude that if a person puts $P$ dollars in an account at an annual interest rate $r,$ compounded continuously, then $A\left(t\right)=P{e}^{rt}.$ This function may be familiar. Since functions involving base $e$ arise often in applications, we call the function $f\left(x\right)={e}^{x}$ the natural exponential function    . Not only is this function interesting because of the definition of the number $e,$ but also, as discussed next, its graph has an important property.

Since $e>1,$ we know ${e}^{x}$ is increasing on $\left(\text{−}\infty ,\infty \right).$ In [link] , we show a graph of $f\left(x\right)={e}^{x}$ along with a tangent line to the graph of at $x=0.$ We give a precise definition of tangent line in the next chapter; but, informally, we say a tangent line to a graph of $f$ at $x=a$ is a line that passes through the point $\left(a,f\left(a\right)\right)$ and has the same “slope” as $f$ at that point $.$ The function $f\left(x\right)={e}^{x}$ is the only exponential function ${b}^{x}$ with tangent line at $x=0$ that has a slope of 1. As we see later in the text, having this property makes the natural exponential function the most simple exponential function to use in many instances.

## Compounding interest

Suppose $\text{}500$ is invested in an account at an annual interest rate of $r=5.5%,$ compounded continuously.

1. Let $t$ denote the number of years after the initial investment and $A\left(t\right)$ denote the amount of money in the account at time $t.$ Find a formula for $A\left(t\right).$
2. Find the amount of money in the account after $10$ years and after $20$ years.
1. If $P$ dollars are invested in an account at an annual interest rate $r,$ compounded continuously, then $A\left(t\right)=P{e}^{rt}.$ Here $P=\text{}500$ and $r=0.055.$ Therefore, $A\left(t\right)=500{e}^{0.055t}.$
2. After $10$ years, the amount of money in the account is
$A\left(10\right)=500{e}^{0.055·10}=500{e}^{0.55}\approx \text{}866.63.$

After $20$ years, the amount of money in the account is
$A\left(20\right)=500{e}^{0.055·20}=500{e}^{1.1}\approx \text{}1,502.08.$

If $\text{}750$ is invested in an account at an annual interest rate of $4%,$ compounded continuously, find a formula for the amount of money in the account after $t$ years. Find the amount of money after $30$ years.

$A\left(t\right)=750{e}^{0.04t}.$ After $30$ years, there will be approximately $\text{}2,490.09.$

## Logarithmic functions

Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry or decibels in sound levels.

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fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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