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6.095 Applications of exponentials

Lesson

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$1aeax+b+C
$\frac{1}{ax+b}$1ax+b $\frac{1}{a}\ln\left(ax+b\right)+C$1aln(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=A0ekt, 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=12A0 and solve for $k$k.

Do:

$\frac{1}{2}A_0$12A0 $=$= $A_0e^{-1200k}$A0e1200k

Substitute $t=1200$t=1200 and $A=\frac{1}{2}A_0$A=12A0 into the given formula.

$\frac{1}{2}$12 $=$= $e^{-1200k}$e1200k

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 $=$= $\ln e^{-1200k}$lne1200k

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}$1000eln21200t

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)eln21200t

Differentiate using $\frac{d}{dx}e^{f\left(x\right)}=f\ '\left(x\right)e^{f\left(x\right)}$ddxef(x)=f (x)ef(x).

  $=$= $-\frac{5\ln2}{6}e^{-\frac{\ln2}{1200}t}$5ln26eln21200t

Simplify.

 

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

$A'\left(400\right)$A(400) $=$= $-\frac{5\ln2}{6}e^{-\frac{\ln2}{1200}\times400}$5ln26eln21200×400

Substitute $t=400$t=400.

  $=$= $-\frac{5\ln2}{6}e^{-\frac{\ln2}{3}}$5ln26eln23

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

$100$100 grams of sugar is placed in a container of water and begins to dissolve.

After $t$t hours, the amount $A$A grams of undissolved sugar remaining is given by $A=100e^{-kt}$A=100ekt.

After $5$5 hours, $6.4$6.4 grams of undissolved sugar remains.

  1. Solve for the exact value of $k$k.

  2. Using the exact value of $k$k found in part (a), solve for the number of hours $t$t it takes for $20$20 grams of sugar to remain undissolved to the nearest two decimal places.

Question 2

The functions $f\left(x\right)=3-e^x$f(x)=3ex and $g\left(x\right)=e^x+1$g(x)=ex+1 intersect at point $A$A.

  1. Solve for the $x$x-coordinate of $A$A. Write each line of working as an equation.

  2. Find the exact value of the area bound between the two curves and the line $x=\ln3$x=ln3.

 

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? 

Graphs of $y=2^x$y=2x$y=e^x$y=ex and $y=3^x$y=3x

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 $\frac{d}{dx}e^{f(x)}=f'\left(x\right)e^{f\left(x\right)}$ddxef(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.

Differentiating $y=a^x$y=ax
$\frac{d}{dx}a^x$ddxax $=$= $\left(\ln a\right)a^x$(lna)ax

 

Worked example

Example 2

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

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

Integrating $y=a^x$y=ax
$\int a^x\ dx$ax dx $=$= $\frac{1}{\ln a}a^x+C$1lnaax+C, where $C$C 

 

Practice questions

Question 3

Consider the function $y=7^x$y=7x.

  1. Using the fact that $e^{\ln a}=a$elna=a, rewrite the function in terms of natural base $e$e.

    Enter each line of work as an equation.

  2. Hence determine $y'$y. Express the derivative in terms of the base $7$7.

    You may use the substitution $u=\left(\ln7\right)x$u=(ln7)x.

  3. Hence determine the exact gradient at $x=1$x=1.

Question 4

A population of wombats increases according to the model $P\left(t\right)=1800\times\left(1.05\right)^t$P(t)=1800×(1.05)t where $t$t is in years.

  1. Express the function in the form $P\left(t\right)=1800e^{kt}$P(t)=1800ekt.

  2. Hence or otherwise, find the rate of change of the population $r$r at $t=3$t=3.

    Round your answer to the nearest whole number.

  3. Solve for the value of $t$t when the rate of change is $150$150 wombats per year.

    Round your answer to the nearest year.

Question 5

Consider the function $f\left(x\right)=2\times3^x$f(x)=2×3x.

  1. Express the function in the form $f\left(x\right)=2e^{kx}$f(x)=2ekx.

  2. Hence, find the exact area bound by $f\left(x\right)$f(x), the $x$x-axis and the lines $x=0$x=0 and $x=1$x=1.

Outcomes

3.2.3.3

use trigonometric functions and their derivatives to solve practical problems; including trigonometric functions of the form 𝑦 = sin(𝑓(𝑥)) and 𝑦 = cos(𝑓(𝑥)).

3.3.3.1

calculate the area under a curve

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