1. Chapter 5: Exponential and Logarithmic Functions 5
Chapter 5: Exponential and Logarithmic Functions 5.6: Solving Exponential Logarithmic Equations Day 1
Essential Question:
Give examples of equations that can be solved by using the properties of exponents and logarithms.

2. 5.6: Solving Exponential and Logarithmic Equations
Powers of the Same Base
Solve the equation 8x = 2x+1
8x = 2x+1
(23)x = 2x+1
23x = 2x+1
Set the exponents equal to each other
3x = x+1
2x = 1
x = 1/2

3. 5.6: Solving Exponential and Logarithmic Equations
Powers of the Different Bases
Solve the equation 5x = 2
5x = 2
log52 = x
log 2/log 5 = x
x =

4. 5.6: Solving Exponential and Logarithmic Equations
Powers of the Different Bases
Solve the equation 24x-1 = 31-x
Take one base and make it into a log problem
log231-x = 4x-1
(1 – x)log23 = 4x-1
(1 – x)(log 3/log 2) = 4x – 1
(1 – x)(1.5850) = 4x – 1 Calculate log 3/log 2
– x = 4x – 1 Distribute on left
– x = 4x Add 1 to both sides
= x Add x to both sides
x = Divide by

5. 5.6: Solving Exponential and Logarithmic Equations
Using Substitution
Solve the equation ex – e-x = 4
ex – e-x = 4
Multiply all terms by ex to remove the negative exponent
e2x – 1 = 4ex
Set everything equal to 0, substitute u = ex
e2x – 4ex – 1 = 0
u2 – 4u – 1 = 0 This is now a…
Quadratic Equation

6. 5.6: Solving Exponential and Logarithmic Equations
Using Substitution
Set u back to ex, and solve

7. 5.6: Solving Exponential and Logarithmic Equations
Assignment
Page 386
Problems 1-31, odd problems
Show work

8. Chapter 5: Exponential and Logarithmic Functions 5
Chapter 5: Exponential and Logarithmic Functions 5.6: Solving Exponential Logarithmic Equations Day 2
Essential Question:
Give examples of equations that can be solved by using the properties of exponents and logarithms.

9. 5.6: Solving Exponential and Logarithmic Equations
Applications of Exponential Equations
Radiocarbon Dating
The half-life of carbon-14 is 5730 years, so the amount of carbon-14 remaining at time t is given by
Many of these problems will deal with percentage of carbon-14 remaining, so P = 1 (i.e. 100%), and the amount remaining will be the percentage left.

10. 5.6: Solving Exponential and Logarithmic Equations
Applications: Carbon Dating
The skeleton of a mastodon has lost 58% of its original carbon-14. When did the mastodon die?
If 58% has been lost, then 42% remains

11. 5.6: Solving Exponential and Logarithmic Equations
Applications: Compound Interest
If $3000 is to be invested at 8% per year, compounded quarterly, in how many years will the investment be wroth $10,680?

12. 5.6: Solving Exponential and Logarithmic Equations
Assignment
Page 386
Problems 53-67, odd problems
Show work

13. Chapter 5: Exponential and Logarithmic Functions 5
Chapter 5: Exponential and Logarithmic Functions 5.6: Solving Exponential Logarithmic Equations Day 3
Essential Question:
Give examples of equations that can be solved by using the properties of exponents and logarithms.

14. 5.6: Solving Exponential and Logarithmic Equations
Applications: Population Growth
A culture started at 1000 bacteria. 7 hours later, there are 5000 bacteria. Find the function and when there are 1 billion bacteria.
Function is based off A = Pert. Need to find r.

15. 5.6: Solving Exponential and Logarithmic Equations
Applications: Population Growth
To find A=1,000,000, need to find t

16. 5.6: Solving Exponential and Logarithmic Equations
Solve the equation ln(x – 3) + ln(2x + 1) = 2(ln x)
ln[(x – 3)(2x + 1)] = ln x2
ln(2x2 – 5x – 3) = ln x2
Natural logs cancel each other out
2x2 – 5x – 3 = x2
x2 – 5x – 3 = 0
Use quadratic equation

17. 5.6: Solving Exponential and Logarithmic Equations
Solve the equation ln(x – 3) + ln(2x + 1) = 2(ln x)
Because = , it’s undefined for ln(x – 3), so there’s only one solution

18. 5.6: Solving Exponential and Logarithmic Equations
Equations with logarithmic & constant terms
Solve ln(x – 3) = 5 – ln(x – 3)
ln(x – 3) + ln(x – 3) = 5
2 ln(x – 3) = 5
ln (x – 3) = 2.5
e2.5 = x – 3
e = x
x =

19. 5.6: Solving Exponential and Logarithmic Equations
Equations with logarithmic & constant terms
Solve log(x – 16) = 2 – log(x – 1)
log(x – 16) + log(x – 1) = 2
log [(x – 16)(x – 1)] = 2
log (x2 – 17x + 16) = 2
102 = x2 – 17x + 16
0 = x2 – 17x – 84
0 = (x – 21)(x + 4)
x = 21 or x = -4
x = -4 would give log(-4 – 16) = log -20, which is undefined
There is only one solution, x = 21

20. 5.6: Solving Exponential and Logarithmic Equations
Assignment
Page 386
Problems & 69-75, odd problems
Show work