1. MAT 150 Module 9 – Exponential and Logarithmic Functions
Lesson 3 – Exponential and Logarithmic Equations
Welcome to Module 9 Lesson 3 - Exponential and Logarithmic Equations

2. Exponential vs. Logarithmic equations
Exponential equations are equations where the variable we want to solve for is in the exponent.
Logarithmic equations are equations where the variable we want to solve for is in a logarithm.
Exponential equations are equations where the variable we want to solve for is in the exponent. For example, we might be asked to solve 4e2x =81. The variable, x, is in the exponent, so this is an exponential equation. Logarithmic equations are equations where the variable we want to solve for is in a logarithm.
For example, log4x – log4(x-1) = ½ is a logarithmic equation because the variable we want to solve for is inside a logarithm with base 4.

3. Example 1 – Exponential Equations
Solve for x: a. 8x + 4 = 10 b. 4e(2x+1) + 2 = 13
Here are some examples of exponential equations. Write these down and we will discuss the procedure for solving them.

4. Exponential Equations
For exponential equations, when the variable is in an exponent, use the following procedure to solve.
Get aN or eN by itself.
N represents the entire exponent.
Then
ln aN = N*ln a
If you have aN take the natural log (base e) of both sides.
For exponential equations, when the variable is in an exponent, use the following procedure to solve:
N represents the entire exponent.
Get aN or eN by itself.
If you have aN take the natural log (base e) of both sides. Then our log rules tell us that
ln aN = Nln(a).
If you have eN take the natural log (base e) of both sides. Then our log rules tell us that
ln eN = (N)ln e = N, since ln e = 1
Solve for x.
Then
ln eN
= N*ln e
= N
If you have eN take the natural log (base e) of both sides.

5. Example 1 - Solution Solve for x: a. 8x + 4 = 10 First, get 8x
by itself by subtracting 4 from both sides.
8x= 6
Take the natural log of both sides.
Ln(8x) = Ln(6)
8x + 4 = 10
First, get 8x by itself by subtracting 4 from both sides. 8x= 6. Next, take the natural log of both sides. Then we have Ln(8x) = Ln(6). Use the power rule to bring down the exponent, x, on the left side. So we have X*Ln(8) = Ln(6). Then divide both sides by Ln(8) to get x by itself. So x = Ln(6)/Ln(8) ≈
Use the power rule.
X*Ln(8) = Ln(6)
Divide both sides by Ln(8).
X= ≈

6. Example 1 - Solution Solve for x: b. 4e(2x+1) + 2 = 13
First, get 4e(2x+1)
by itself by subtracting 2 from both sides.
4e(2x+1) = 11
e(2x+1) =
Divide both sides by 4 to get by e(2x+1 ) by itself.
Take the natural log of both sides.
Ln(e(2x+1)) = Ln( )
Solve for x:
b. 4e(2x+1) + 2 = 13
First, get 4e(2x+1) by itself by subtracting 2 from both sides. Then we have 4e(2x+1) = 11. Next, divide both sides by 4 to get by e(2x+1) itself. That gives us e(2x+1) = 11/4. Then take the natural log of both sides. So Ln(e(2x+1)) = Ln(11/4). Use the power rule, to bring down the power, 2x+1. So we have (2x+1)Ln(e) =Ln(11/4). Ln(e) = 1 so this simplifies to 2x+1 = Ln(11/4). Finally, get x by itself by subtracting 1 and dividing by 2 from both sides. Then we have x = [Ln(11/4)-1]/2 ≈
Use the power rule.
(2x+1)Ln(e) = Ln( )
2x+1 = Ln( )
Get x by itself by subtracting 1 and dividing by 2.
X= Ln −1 2 ≈

7. Solving Exponential Equations Graphically
Graph the left side of the equation as Y1 and the right side of the equation as Y2. Set the viewing window so that you can see the point of intersection of the two lines. The x coordinate of the point of intersection is the solution to the equation!
To solve exponential equations graphically:
Graph the left side of the equation as Y1 and the right side of the equation as Y2.
Set the viewing window so that you can see the point of intersection of the two lines.
The x coordinate of the point of intersection is the solution to the equation!

8. Example 1 – Graphical Solution
Here you can see the graphical solution to Part a of example 1. The solution we found was or

9. Example 1 – Graphical Solution
Here you can see the graphical solution to Part b of example 1. The solution we found was Note that I had to put parentheses around the exponent in order to get the desired result.

10. Application of Exponential Functions – Example 2
The formula y = 250e x gives the amount of a 250 gram radioactive sample that will be left after x years.
How many grams are left after 100 years?
Graph the function on the window [0, 10,000] by [0, 250].
What is the half life of the sample?
The formula y = 250e x gives the amount of a 250 gram radioactive sample that will be left after x years.
How many grams are left after 100 years?
Graph the function on the window [0, 10,000] by [0, 250].
What is the half life of the sample?

11. Application of Exponential Functions – Example 2 Solution
How many grams are left after 100 years? Round your answer to the nearest tenth.
To find the amount left after 100 years, plug in x = 100.
How many grams are left after 100 years? Round your answer to the nearest tenth.
To find the amount left after 100 years, plug in x = 100.
y = 250e( (100))
= grams.
y = 250e( (100))
= grams.

12. Application of Exponential Functions – Example 2 Solution
b. Graph the function on the window [0, 10,000] by [0, 250].
Justin – can you make it pop out?

13. Application of Exponential Functions – Example 2 Solution
c. What is the half life of the sample?
The half life refers to the number of years it will take for half the original amount to remain.
Since there are 250 grams originally, half that amount will be 125 grams.
c. What is the half life of the sample?
The half life refers to the number of years it will take for half the original amount to remain. Since there are 250 grams originally half that amount will be 125 grams.
So we want to solve the equation 250e x = 125 for x. Since we have already graphed the function, let’s use graphical methods to solve.
Solve the equation
250e x = 125
for x.

14. Application of Exponential Functions – Example 2 Solution
Using the graph to solve, when x=2708 there are 125 grams remaining. So the half life is approximately 2,708 years. After 2708 years there will be half the original amount remaining.

15. Solving Logarithmic Equations – Example 3
Solve the logarithmic equations for x:
(1/3)Ln (2x+3) + 7 = 11
5 = 1 + log3t + log3 (t+6)
log2(t + 4) = 1 + log2(5t)
Now we want to solve some equations involving logarithms. Write down these examples and we will discuss the procedure for solving.

16. Logarithmic Equations
For logarithmic equations, use the following procedure to solve:
If you have more than one log, combine them into one using log rules.
N represents whatever we are taking the log of.
Get loga(N) by itself.
For logarithmic equations, use the following procedure to solve:
N represents whatever we are taking the log of.
If you have more than one log, combine them into one using log rules.
Get loga(N) by itself.
Rewrite loga N = M in exponential form aM = N
aM = N
Rewrite loga N = M in exponential form:
Solve for x.

17. Solving Logarithmic Equations – Example 3
Solve the logarithmic equation for x:
(1/3)Ln (2x+3) + 7 = 11
Subtract 7 from both sides
(1/3)Ln(2x+3) = 4
First get Ln(2x+3) by itself
Multiply both sides by 3
Ln(2x+3) = 12
Rewrite in exponential form
e12 = 2x+3
Solve the logarithmic equations for x:
(1/3)Ln (2x+3) + 7 = 11
First get Ln(2x+3) by itself by subtracting 7 from both sides and multiplying both sides by 3. Then rewrite in exponential form. Ln(2x+3) has base e, so it can be rewritten as e12 = 2x+3. Then solve for x by subtracting 3 from both sides and dividing both sides by 2. So x = (e^12 – 3)/2 ≈ or
e = 2x
Subtract 3 from both sides
Solve for x
Divide both sides by 2
x = ≈ 81,

18. Solving Logarithmic Equations – Example 3
Solve the logarithmic equations for t: 5 = 1 + log3t + log3 (t+6)
First get the logarithms by themselves
Subtract 1 from both sides
4 = log3t + log3 (t+6)
Rewrite the two logarithms as one using the addition rule
4 = log3[(t)(t+6)]
4 = log3(t2 + 6t)
Rewrite in exponential form
To solve the equation in part b, first get the logarithms by themselves by subtracting 1 from both sides. So we have 4 = log3t + log3 (t+6). Then use the addition rule to rewrite the two logarithms as one. log3t + log3 (t+6) = log3[(t)(t+6)], then we can multiply t(t+6) to get log3(t2 + 6t). Next rewrite in exponential form. 3 is the base, 4 is the exponent, and t2 + 6t is the output, so in exponential form we have 34 = t2 + 6t which we can simplify to 81 = t2 + 6t. Now this is a quadratic equation, so we need to use quadratic methods to solve, like factoring or the quadratic formula. We also need to move the 81 over so we have t2 + 6t - 81= 0.
Using the quadratic formula, we end up with t = -3+3√10 ≈ and t = √10 ≈ However, the logarithmic function is always positive, so looking back at our original equation, log3t would be undefined for t ≈ So we toss out that solution and our solution is t ≈
34 = t2 + 6t
81 = t2 + 6t
t =
t ≈
t ≈
Solve for t:
t2 + 6t – 81 = 0

19. Solving Logarithmic Equations – Example 3
Solve the logarithmic equations for t: log2(t + 4) = 1 + log2(5t)
Move log2(5t) to the left by subtracting it
log2(t + 4) - log2(5t) = 1
Combine the two logarithms into one using the subtraction rule
Solve the logarithmic equations for t: log2(t + 4) = 1 + log2(5t).
First move the two logarithms to the left by subtracting log2(5t) from both side, so we have log2(t + 4) - log2(5t) = 1. Next combine the two logarithms into one using the subtraction rule. log2(t + 4) - log2(5t) becomes log2[(t + 4)/(5t)] .
Next rewrite in exponential form. 2 is the base, 1 is the exponent, and (t+4)/(5t) is the output, so we have 2^1 =2 = (t+4)/(5t). To solve for t, we multiply both sides by 5t to get 2(5t) = t+4 or 10t = t+4. Then we subtract t from both sides, which gives us 9t = 4. Finally we divide both sides by 9 to get t = 4/9 which is 0.4 repeating.
Rewrite in exponential form
Solve for t
2(5t) = t + 4
10t = t + 4
9t = 4
t = 4/9 = …

20. Solving Logarithmic Equations Graphically
We can solve logarithmic equations the same way we solved exponential equations using the graphing calculator. Just graph the left side as y1 and the right side as y2, and find the point of intersection. The solution is the x value of the point of intersection.
We can also confirm our solutions to these three equations graphically. We can solve logarithmic equations the same way we solved exponential equations using the graphing calculator. Just graph the left side as y1 and the right side as y2, and find the point of intersection. The solution is the x value of the point of intersection.

21. Solving Logarithmic Equations Graphically
Here is the graphical solution to the first equation. The solution we found by hand was x = 81,375.9.
Note that I had to do some manual adjustment to the window in order to increase the x axis while keeping the y axis near y = 11.

22. Solving Logarithmic Equations Graphically
Here is the graphical solution to the second equation. Note that I changed the variable t to x when I typed the equation. Also, the logarithmic functions can be found in the “functions” menu in Desmos under “misc”.

23. Solving Logarithmic Equations Graphically
Here is the graphical solution to the third equation. Note that I changed the variable t to x again. The solution we found was 4/9 which is 0.4 repeating.

24. Next up:
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Module 10