6.095 Applications of exponentials

Using our knowledge of the natural logarithm function and its inverse relationship with the exponential function, we can apply our newly acquired tools in solving equations and calculus to a wide range of applications. We can now find algebraic solutions for exponential problems where previously we required technology, as well as being able to apply calculus to general exponential functions. 

Keep the following rules in mind as we look at some examples.

Function	Derivative
$e^{f\left(x\right)}$ef(x)	$f\ '\left(x\right)e^{f\left(x\right)}$f ′(x)ef(x)
$\ln\left(f\left(x\right)\right)$ln(f(x))	$\frac{f\ '\left(x\right)}{f\left(x\right)}$f ′(x)f(x)​

 

Function $f\left(x\right)$f(x)	Integral $\int f\left(x\right)dx$∫f(x)dx
$e^{ax+b}$eax+b	$\frac{1}{a}e^{ax+b}+C$1a​eax+b+C
$\frac{1}{ax+b}$1ax+b​	$\frac{1}{a}\ln\left(ax+b\right)+C$1a​ln(ax+b)+C, for $ax+b>0$ax+b>0

 

Worked examples

Example 1

A sample of a radioactive substance decays exponentially according to the model $A=A_0e^{-kt}$A=A0​e−kt, where $A$A is the amount of the substance remaining after $t$t days.

(a) Solve for the exact value of $k$k given the half-life of the substance is $1200$1200 days.

Think: We know when $t=1200$t=1200 there will be half the initial amount left. Substitute $t=1200$t=1200 and $A=\frac{1}{2}A_0$A=12​A0​ and solve for $k$k.

Do:

$\frac{1}{2}A_0$12​A0​	$=$=	$A_0e^{-1200k}$A0​e−1200k	Substitute $t=1200$t=1200 and $A=\frac{1}{2}A_0$A=12​A0​ into the given formula.
$\frac{1}{2}$12​	$=$=	$e^{-1200k}$e−1200k	Divide by $A_0$A0​, the resulting equation shows we are solving for when the multiplier is $\frac{1}{2}$12​.
$\ln\frac{1}{2}$ln12​	$=$=	$\ln e^{-1200k}$lne−1200k	Take the natural logarithm of both sides.
$-\ln2$−ln2	$=$=	$-1200k$−1200k	Simplify both sides using logarithm laws.
$\therefore k$∴k	$=$=	$\frac{\ln2}{1200}$ln21200​	Solve for $k$k.

 

(b) If there is initially $1000$1000 grams of the substance, at what rate is the substance decaying at $t=400$t=400 days?

Think: Use differentiation to find the rate of change of the substance at any time $t$t, then substitute $t=400$t=400. That is find $A'\left(400\right)$A′(400).

Do: Find $A'\left(t\right)$A′(t):

$A\left(t\right)$A(t)	$=$=	$1000e^{-\frac{\ln2}{1200}t}$1000e−ln21200​t	Substitute the initial mass given and the value of $k$k found in part (a)
$\therefore\ A'\left(t\right)$∴ A′(t)	$=$=	$1000\times\left(-\frac{\ln2}{1200}\right)e^{-\frac{\ln2}{1200}t}$1000×(−ln21200​)e−ln21200​t	Differentiate using $\frac{d}{dx}e^{f\left(x\right)}=f\ '\left(x\right)e^{f\left(x\right)}$ddx​ef(x)=f ′(x)ef(x).
	$=$=	$-\frac{5\ln2}{6}e^{-\frac{\ln2}{1200}t}$−5ln26​e−ln21200​t	Simplify.

 

Hence, find $A'\left(400\right)$A′(400):

$A'\left(400\right)$A′(400)	$=$=	$-\frac{5\ln2}{6}e^{-\frac{\ln2}{1200}\times400}$−5ln26​e−ln21200​×400	Substitute $t=400$t=400.
	$=$=	$-\frac{5\ln2}{6}e^{-\frac{\ln2}{3}}$−5ln26​e−ln23​	Simplify.
	$\approx$≈	$-0.458$−0.458	Evaluate.

 

Hence, at $t=400$t=400 the substance is decaying at a rate of approximately $0.458$0.458 g/day (don't forget units).

 

Practice questions

Question 1

Question 2

 

Calculus and the general exponential function $y=a^x$y=ax

We have seen how to differentiate the exponential function $y=e^x$y=ex but how can we differentiate an exponential function with a different base? 

The three functions above can be viewed as horizontal dilations of each other. Using transformations we can in fact write any exponential function with a base of $e$e. This implies any function of the type $y=a^x$y=ax can be written in the form $y=e^{kx}$y=ekx, the derivative can then be found using our prior knowledge $\frac{dy}{dx}=ke^{kx}=ka^x$dydx​=kekx=kax. So the derivative of any exponential function of the form $y=a^x$y=ax is simply a constant multiple of itself, with $a=e$a=e being the special case where the multiple is $1$1. We investigated this fact when looking to find the derivative of $y=e^x$y=ex.

 

So, how do we write $a^x$ax with a base $e$e? We can use the fact that $e^{\ln a}=a$elna=a. Recall, by definition $\ln a$lna is the index if $e$e is raised to results in $a$a.

Thus,

$y$y	$=$=	$a^x$ax
	$=$=	$\left(e^{\ln a}\right)^x$(elna)x
	$=$=	$e^{x\ln a}$exlna

 

Taking the derivative of both sides using $\frac{d}{dx}e^{f(x)}=f'\left(x\right)e^{f\left(x\right)}$ddx​ef(x)=f′(x)ef(x):

$\frac{dy}{dx}$dydx​

$=$=

$\left(\ln a\right)e^{x\ln a}$(lna)exlna

 

But, remember $y=e^{x\ln a}=a^x$y=exlna=ax, therefore:

$\frac{dy}{dx}$dydx​

$=$=

$\left(\ln a\right)a^x$(lna)ax

 

Therefore, the the derivative of an exponential function is the product of the original function and the constant $\ln a$lna. This agrees with the special case of the exponential function $y=e^x$y=ex, as $\ln e=1$lne=1.

Note: Brackets have been placed around the constant term $\ln a$lna to make it clear that the term $a^x$ax is not part of the argument of the logarithm term. An alternative is to place the constant term second as $a^x\ln a$axlna.

 

Worked example

Example 2

Differentiate the function $y=2^x$y=2x.

Think:  This is an exponential function with $a\ne e$a≠e. Therefore we will use $\frac{d}{dx}a^x=\left(\ln a\right)a^x$ddx​ax=(lna)ax.

Do:  

$y$y	$=$=	$2^x$2x
$y'$y′	$=$=	$\left(\ln2\right)2^x$(ln2)2x

 

Using the same method to rewrite an exponential function of the form $y=a^x$y=ax as $y=e^{x\ln a}$y=exlna, also allows us to perform integration for general exponential functions.

 

Practice questions

Question 3

Question 4

Question 5