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The number $e,$ commonly called the natural base, is a mathematical constant named by the mathematician Leonhard Euler.

$e = 2.7182818284\ldots$

The value of the number $e$ can be found in different ways. Here, compound interest will be used to find its value. The formula for compound interest is the exponential growth function $y=P\left(1+\frac{r}{n}\right)^{nt},$ where the constant $r$ is the interest rate in decimal form. If the interest rate was $100\,\%,$ which is extremely profitable, the value of $r$ equals $1.$ $y=P\left(1+\frac{1}{n}\right)^{nt}.$ The number $n$ is the amount of times the interest is compounded each year. That is, how often the accrued interest is added to the balance. The more often the interest is compounded, the higher the profit will be each year. What happens if the interest is compounded very often? That is, when $n$ is large. To examine this, the function will be rewritten using the the power of a power property. $y=P\left(1+\frac{1}{n}\right)^{nt}=P\left(\left(1+\frac{1}{n}\right)^n\right)^t$ For the rewritten function, the expression inside the outer parentheses is a constant depending only on $n.$ To analyze what happens when $n$ increases, larger and larger values will be substituted into the expression.

$n$ | $\left(1+\dfrac{1}{n}\right)^n$ | Expression value |
---|---|---|

$10$ | $\left(1+\dfrac{1}{10}\right)^{10}$ | $2.59374\ldots$ |

$100$ | $\left(1+\dfrac{1}{100}\right)^{100}$ | $2.70481\ldots$ |

$1000$ | $\left(1+\dfrac{1}{1000}\right)^{1000}$ | $2.71692\ldots$ |

$10\,000$ | $\left(1+\dfrac{1}{10\,000}\right)^{10\,000}$ | $2.71814\ldots$ |

$100\,000$ | $\left(1+\dfrac{1}{100\,000}\right)^{100\,000}$ | $2.71826\ldots$ |

The table shows that for high values of $n,$ the expression seems to approach $\sim2.718.$ In fact, as $n\rightarrow \infty,$ the value of the expression approaches the natural base $e$: $e=2.71828\ldots$ Substituting $e$ for the expression gives the function $y=Pe^{t},$

which applies when the interest rate is $100\,\%,$ compounded infinitely often. When interest is compounded infinitely often, it is said to beExponential functions are commonly expressed using the natural base $e$, as they then exhibit useful characteristics. These functions are called natural base exponential functions and are written in the form $f(x) = ae^{rx}.$ If $a$ and $r$ are both positive, the function is an exponential growth function. If $a$ is positive and $r$ is negative, the resulting function is instead an exponential decay function.

Determine whether the exponential function
$f(x) = 0.5 e^{0.3x}$
shows growth or decay. Then, rewrite the function in the form $f(x) = ab^x$ and graph it.

The function is currently expressed as a natural base exponential function, $f(x) = ae^{rx}.$ Thus, identifying the signs of the constants $a$ and $r$ will help us decide whether it is an exponential growth or decay function. Both $a$ and $r$ are positive in this case. Thus, it models an exponential growth. To rewrite the function as $f(x) = ab^x,$ we have to make sure that the exponent is nothing but $x.$ This is achieved using the power of a power property and calculating the new base.

$f(x) = 0.5e^{0.3x}$

$f(x) = 0.5\left(e^{0.3}\right)^x$

$f(x) = 0.5 \cdot (1.34985\ldots)^x$

$f(x) \approx 0.5 \cdot 1.35^x$

Now that the function is rewritten, we can graph it as usual, starting by plotting the initial value, $0.5.$

Next, more points are found by repeatedly increasing the $x$-value by $1$ and multiplying the function value by the constant multiplier, $1.35.$

Connecting the points with a smooth curve gives us the desired graph.

When an interest rate of $100\,\%$ is compounded continuously, meaning it is compounded infinitely often, the resulting function is $y = Pe^t.$ When the interest rate is something other than $100,$ the arbitrary rate $r$ is used. $y = P \left( 1 + \dfrac r n \right)^{nt}.$ Suppose that the interest rate doubles, $r = 2.$ Because interest is compounded continuously, this effectively leads to the same growth in half the time. Similarly, three times the interest leads to the same growth in a third of the time. This corresponds to a horizontal stretch or shrink, leading to the following function.

$y = Pe^{rt}$

Carlos won the lottery! In total, his winnings are $\$50\,000.$ He wants to buy a high quality miniature blimp, which costs $\$66\,499.$ Since his winnings aren't enough, he's decided to deposit them into a savings account for $5$ years, earning $6\,\%$ yearly interest compounded continuously. Assuming the price stays constant, will he have enough money in his account to buy the miniature blimp after the $5$ years?

As the interest of the account is continuously compounded, the balance can be modeled with the function $y = Pe^{rt},$ where $P$ is the principal and $r$ is the interest rate in decimal form. In this case, the principal is $\$ 50\,000$ and the interest rate is $0.06,$ resulting in the function $y = 50\,000e^{0.06t}.$ By substituting $t = 5$ into the function, we'll find the balance at $5$ years.

$y = 50\,000e^{0.06t}$

$y = 50\,000e^{0.06 \cdot {\color{#0000FF}{5}}}$

$y = 50\,000e^{0.3}$

$y = 67\,492.94038\ldots$

$y \approx 67\,493$

At $5$ years, his account balance will be $\$ 67\,493.$ Thus, he will be able to afford the miniature blimp at $\$ 66\,499.$

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