Class 10

Between Class Sessions - Prep for Day 10

Please spend around 2 hours working between class sessions, focusing on the tasks below. Use any extra time to complete KnewtonAlta assignments and/or work on Project tasks.

Pick something from your prep today that you can share with your group in class. It might be something new that you learned. It might be questions you have that are still unanswered, or a question along with what helped you eventually answer it. It might be something tricky that you solved. It might be a review topic that helped you remember something. It might be a conversation you had with AI that was helpful. You will have a chance to share this with your peers during class. Come ready to articulate your thinking and questions.

Preparation

(1) Solving Exponential and Log Equation (videos)

Watch these video from mathispower4u.

• Exponential Equations (video 1) (~5 minutes)
• Exponential Equations (video 2) (~4.5 minutes)
• Exponential Equations (video 3) (~3.5 minutes)
• Log Equations (video 1) (~7.5 minutes)
• Log Equations (video 2) (~7.5 minutes)

(2) Transformation (Exercises)

Consider the general form \(P(x;a,b,h,k) = af(c(x-h))+k\) for the transformation of a function \(f(u)\).

• For each \(P(x)\) and parent function \(f(u)\) below, identify the parameters \(a\), \(c\), \(h\), and \(k\) required to obtain \(P(x) = P(x;a,b,h,k)\) and describe how these parameters transform \(f(u)\).
• State the domain and range for each function below.
• Graph each function below.
  • \(P(x) = \ln{(x-2)}\) where \(f(u) = \ln{u} = \log_{e}{u}\)
  • \(P(x) = -3\log{(x)} + 1\) where \(f(u) = \log{u} = \log_{10}{u}\)
  • \(P(x) = -\log_7{(x)} -5\) where \(f(u) = \log_7{u}\)
  • \(P(x) = \ln(6-2x)\) where \(f(u) = \ln{u}\)

Use this code to define the functions above.

p1 <- function(x){ log(x-2) }
p2 <- function(x){ -3*log(x,10) + 1 }
p3 <- function(x){ -log(x,7) - 5 }
p4 <- function(x){ log(6-2*x) }

Use (and adapt) this code to plot the functions above.

x <- seq(-10,10,0.1)
y <- p1(x)

par(mar=c(2.5,2.5,1,0.25))
plot(x,y,type="l",xlim=c(-10,10),ylim=c(-10,10))
abline(h=0,col='gray',lty=3) #x-axis
abline(v=0,col='gray',lty=3) #y-axis
abline(v=2,col='red',lty=3) #vertical asymptote of function

(3) Function Fairy Tales (Shoe Tread)

The tread on the bottom of a boot has been wearing off at a fairly constant rate starting from the date it was purchased. The shoe began with about 60mm of tread, and every month has been loosing about 3mm. A model for the tread thickness is \(f(x;m,b) = mx+b\), where \(x\) is the number of months since the date it was purchased.

For each choice of parameters \(m\) and \(b\) below, tell the story given by the model (even if said story is ludicrous). Then identify the model that best fits the story provided above.

1. \(m=2,b=3\)
2. \(m=0,b=60\)
3. \(m=60,b=0\)
4. \(m=-1,b=10\)
5. \(m=-3,b=60\)
6. \(m=3,b=-60\)

Regular Reminders

Skill Practice (KA Homework)

• Complete 1 – Definition of Logarithm
• Complete 1 – Evaluate Exponential and Logarithmic Functions
• Complete 1 – Properties of Logarithms
• Work on 1 – Solve Logarithmic Equations (and IVT) assignment

Applied Practice (Project Work)

• Continue working on Project 1 Task 3

During Class

Brain Gains

• Write in expanded form.
  • \(\sum_{i=5}^{10} 3i^2\)
  • \(\prod_{k=1}^{4}(100-k^2)\)
• Use log properties to write \(\ln\left(\prod_{k=1}^{4}(100-k^2)\right)\) as a sum of logarithms.
• Solve \(2^x = 37\)
• Solve \(\log(x)+\log(x-1) = \log(6)\)

Group Meeting

Start by giving each person a moment to share what they chose to prepare for class. Help each other address any questions. When each person has had a chance to share, move on the other activities.

Activity - Solving Equations

Solve the following equations for \(x\). Write down the exact answer and decimal approximation (round to 4 decimal places).

1. \(3^x = 17\)

2. \(\ln(4x-2) = 5\)

3. \(\log(5x-30) = 1\)

4. \(9^x = 49\)

5. \(2^x = 56\)

6. \(-2\log_5(7x) = 2\)

7. \(\log_3(x-3) - 4 = 12\)

8. \(\log_2(x-1) + \log_2(x+5) = 4\)

9. \(e^{x-1}-5 = 5\)

10. \(3(4^{2-5x})-7=8\)

11. \(\ln(x+5) + \ln(x) = \ln(12x)\)

Solutions

1. \(3^x = 17 \Rightarrow\) \[\begin{align*} x &= \log_3(17) \\ &= \dfrac{\ln 17}{\ln 3} \\ &\approx 2.5789 \end{align*}\]

2. \(\ln(4x-2) = 5 \Rightarrow 4x-2 = e^5 \Rightarrow\) \[\begin{align*} x &= \dfrac{e^5+2}{4} \\ &\approx 37.6033 \end{align*}\]

3. \(\log(5x-30) = 1 \Rightarrow 5x-30 = 10 \Rightarrow x = 8\)

4. \(9^x = 49 \Rightarrow\) \[\begin{align*} x &= \log_9(49) \\ &= \dfrac{\ln 49}{\ln 9} = \dfrac{\ln 7}{\ln 3} \\ &\approx 1.7712 \end{align*}\]

5. \(2^x = 56 \Rightarrow\) \[\begin{align*} x &= \log_2(56) \\ &= \dfrac{\ln 56}{\ln 2} \\ &\approx 5.8074 \end{align*}\]

6. \(-2\log_5(7x) = 2 \Rightarrow \log_5(7x) = -1 \Rightarrow 7x = \dfrac{1}{5} \Rightarrow x = \dfrac{1}{35} \approx 0.0286\)

7. \(\log_3(x-3) - 4 = 12 \Rightarrow \log_3(x-3) = 16 \Rightarrow\) \[\begin{align*} x &= 3^{16}+3 \\ &= 43{,}046{,}724 \end{align*}\]

8. \(\log_2(x-1)+\log_2(x+5) = 4 \Rightarrow\) \[\begin{align*} (x-1)(x+5) &= 16 \\ x^2+4x-21 &= 0 \\ x &= 3 \quad \text{(reject } x=-7 \text{, domain)} \end{align*}\]

9. \(e^{x-1}-5=5 \Rightarrow e^{x-1}=10 \Rightarrow\) \[\begin{align*} x &= 1+\ln(10) \\ &\approx 3.3026 \end{align*}\]

10. \(3(4^{2-5x})-7=8 \Rightarrow 4^{2-5x}=5 \Rightarrow\) \[\begin{align*} 2-5x &= \log_4(5) \\ x &= \dfrac{2-\log_4(5)}{5} \\ &\approx 0.1678 \end{align*}\]

11. \(\ln(x+5)+\ln(x)=\ln(12x) \Rightarrow x(x+5)=12x \Rightarrow\) \[\begin{align*} x^2-7x &= 0 \\ x &= 7 \quad \text{(reject } x=0 \text{, domain)} \end{align*}\]

Activity - Solving Exponential and Log Equations using Uniroot

Let’s now solve some equations in R using uniroot. To solve the equation \(3^b=17\) with uniroot, the key is to create a function whose zero is the solution we want. So we rewrite this as \(f(b) = 3^b-17\), and we want to solve \(3^b-17=0\). From there, we must give an interval in which we believe the root lies, and then uniroot will find the zero. Because \(3^2 = 9\) and \(3^3=27\), we know the zero is in the interval \((2,3)\). Below is the code for using uniroot to solve \(3^b=17\).

f <- function(x){3^x-17}
interval <- c(2,3)
my_root <- uniroot(f, interval)$root

#The code below produces a plot that illustrates what uniroot found. 
x<-seq(interval[1],interval[2],0.1)
plot(x,f(x),type="l")
abline(h=0,col = "gray")
abline(v=my_root,col = "gray")
points(my_root,0, col = "red")

As a team, try adapting the code above to find the zeros of the problems you solved by hand in the previous activity.

Discussion

Solving equations with uniroot

What does uniroot do?

?uniroot

The key to using uniroot is to pick two \(x\)-values where you know the values of the function differ in sign. If needed, construct a plot to help you pick them. Let’s look at a few code examples.

• Solve \(h(x) = 0\) given \(h(x) = 3x - 15\)

rm(list=ls()) #Clears the environment

h <- function(x){
  3*x -15
}

#What does the $root code do below?
uniroot(h,c(0,30))
uniroot(h,c(0,30))$root

If needed, remember to construct a plot.

x <-seq(0,30,1)
plot(x,h(x), type="l")
abline(h=0, col = "lightgray")

• Solve \(h(x) = 4\) given \(h(x) = 3x - 15\)

h.shift <- function(x){
  h(x) - 4
}

uniroot(h.shift,c(0,5))$root

uniroot(h.shift,c(0,10))$root

• Let’s solve \(3x - 15 = e^{-x+6}\) using uniroot. Note that this function doesn’t have a by-hand solution technique. We’ll have to update the bounds in uniroot below, using the graph.

g <- function(x){
  3*x-15-exp(-x+6)
}

uniroot(g,c(0,5))$root

x <-seq(0,30,1)
plot(x,g(x), type="l")
abline(h=0, col = "lightgray")

• What happens when we try to solve something impossible, such as \(\frac{1}{x} = 0\).

f <- function(x){
  1/x
}

uniroot(f,c(-10,-3))$root

uniroot(f,c(-1,1))$root

uniroot(f,c(-1,1))

Let’s practice using uniroot to solve some of the equations below. Remember, we start by defining the function that you wish to equal zero (you might need to subtract), and then use uniroot. If needed, we can make a plot first to figure out an interval where the zero should appear.

1. \(3x-5=0\)

2. \(3x-5=7\)

3. \(3x-5=e^{-x}\) (Remember to use exp(-x)) for \(e^{-x}\) )

4. \(3x-5=\ln(x)\) (Remember to use log(x) for \(\ln(x)\)).

5. \(x^2+x-6=0\). There are two different solutions. See if you can capture both by choosing different lower and upper limits.

6. \(x^2-8x+12=0\). There are two different solutions. See if you can capture both by choosing different lower and upper limits.

Intermediate Value Theorem

Intermediate Value Theorem (IVT)

If \(f(x)\) is a continuous function on a closed interval \([a, b]\), where \(f(a) \neq f(b)\), and \(y_0\) is any \(y\)-value strictly between \(f(a)\) and \(f(b)\), then \(y_0 = f(x_0)\) for some \(x\)-value, \(x_0\), in \([a, b]\).

Note: While there is a rigorous mathematical definition for what it means to be a continuous function, for our purposes the following concept with suffice, a function is continuous on an interval if the graph of that function can be traced with a pencil without lifting the pencil from the page.

Consider the function \(f(x) = (-13 + 2x)e^{-0.05x}\) and the interval \([3, 50]\).

• What does the IVT say about this function over this interval?

Expand for a plot of the function.

rm(list=ls())
f <- function(x){ (-13 + 2*x)*exp(-0.05*x) }

x <- seq(0,60,0.1)
par(mar=c(2.5,2.5,0.5,0.5))
plot(x, f(x), type='l')
abline(h=0,col='gray',lty=3)
abline(v=0,col='gray',lty=3)
abline(v=3,col='red',lty=3)
abline(v=50,col='red',lty=3)
abline(h=f(3),col='blue',lty=3)
abline(h=f(50),col='blue',lty=3)

• We currently have \(f(x) = (-13 + 2x)e^{-0.05x}\) with \([a, b] = [3, 50]\), which means \(a = 3\) and \(b = 50\).
  
• We can compute know \(f(a) = f(3) \approx -6.024956\) and \(f(b) = f(50) \approx 7.141395\). Note that \(f(3) \neq f(50)\).
• We can pick \(y_0 = 0\) because we know \(f(a)<0<f(b)\) (as \(-6.024956 < 0 < 7.141395\)).
• The intermediate value theorem states that there must exist some \(x\)-value, which we’ll call \(x_0\), so that \(f(x_0) = y_0 = 0\).
• We can solve \((-13 + 2x+0)e^{-0.05x_0}=0\) for \(x_0\) and get an exact solution (or analytic solution).

Expand to see an exact solution to \(f(x_0) = 0\), obtaining \(x_0=6.5\).

We want to solve \((-13 + 2x_0) e^{-0.05x_0} = 0\). Since we have two things that multiply to zero, we know one of the quantities must be zero. We see \(-13 + 2x_0 = 0\) or \(e^{-0.05x_0} = 0\). Since the range of an exponential function is strictly positive numbers, we know \(e^{-0.05x_0} \neq 0\). So we only have to determine when \(-13 + 2x_0 = 0\). This means \(2x_0 = 13\), or \(x_0 = \frac{13}{2} = 6.5\).

The intermediate value theorem does not require that we pick \(y_0 = 0\).

• What values can we choose for \(y_0\)? Let’s choose \(y_0 = 5\), and repeat what we did above.

• Why is \(y_0=5\) a valid choice?

• Let’s try solving \(f(x_0) = y_0 = 5\) by hand for \(x_0\). (Unfortunately we will fail when trying to solve this equation exactly.)

Expand to see an attempt at solving \(f(x_0) = 5\) exactly.

We wish to solve \((-13 + 2x_0) e^{-0.05x_0} = 5\). We want to isolate \(x_0\), so let’s start by expanding the left-hand side (LHS) of the equation. This gives us \[-13e^{-0.05x_0} + 2x_0e^{-0.05x_0} = 5\] That didn’t help much. While all of the \(x_0\)’s are on the LHS, there are no like terms. The equation looks different but we are no closer to our goal of isolating \(x_0\).

Instead, let’s try isolating the exponential and using a logarithm to rewrite the equation. Isolating the exponential term in \((-13 + 2x_0) e^{-0.05x_0} = 5\) gives \[e^{0.05x_0} = \frac{-13 + 2x_0}{5}\] From here we can rewrite in logarithmic form to obtain \(0.05x_0 = \ln \left(\frac{-13 + 2x_0}{5} \right)\), which requires \(-13 + 2x_0 > 0\) as we cannot compute logarithm of a negative number. We can rewrite this as \[x_0 = \frac{1}{0.05}\ln \left(\frac{-13 + 2x_0}{5} \right),\] but this doesn’t provide a solution as we still have an \(x_0\) trapped in a logarithm. This approach failed as well.

It turns out that there is no known solution technique for finding an exact solution to this type of problem. We’ve already entered a realm where current knowledge has failed to find exact solutions.

• While we cannot obtain an exact solution to \((-13 + 2x_0) e^{-0.05x_0} = 5\), let’s solve numerically using R.
• Remember that to use the uniroot() function, we need to have one side of the equation be zero (we need to be looking for a root). We see \(f(x_0) = 5\) is equivalent to \(f(x_0) - 5 = 0\). The function \(f(x_0) - 5\) is a new function, let’s call it \(g\). Notice that \(g\) is a transformation of \(f\). To get \(g\) from \(f\) we just shift down 5.

Expand for the solution to \(f(x_0) = 5\) using uniroot.

# I may call on people to share in class.
par(mar=c(2.5,2.5,0.5,0.5))
plot(x, f(x), type='l',col='gray')
abline(h=0,col='gray',lty=3)
abline(v=0,col='gray',lty=3)

a <- 3
b <- 50
y0 <- 5
g <- function(x,y0){f(x)-y0}
x0 <- uniroot(g,c(a,b),y0=y0)$root
lines(x, g(x,y0))
points(c(a,b,a,b), c(f(a),f(b),g(a,y0),g(b,y0)),pch=16,col=2)
points(x0,f(x0),pch=16,col='blue')
points(x0,g(x0,y0),pch=16,col='blue')

x0

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Source: Class.10 on byuimath.com