1. Warm-Up Solve the following equations: 5 𝑥 =337 log 10=𝑥 log 6 𝑥 =1
65 𝑥 =3409

2. Properties of Logs Pt. 2
Section 7.2.2

3. Learning Targets Determine the Product Property of Logs
How to use the Product Property of Logs
Determine the Quotient Property of Logs
How to use the Quotient Property of Logs

4. Recap Power Property of Logs
log 𝑏 𝑥 =𝑥 log 𝑏
Allows us to quickly solve for unknown values that are exponents

5. Figuring out the Log Property
Use your calculator to solve for x below:
log log 6 = log 𝑥
log log 2 = log 𝑥
log log = log 𝑥
log log 17 = log 𝑥
log log 𝜋 = log 𝑥
log 𝑎 + log 𝑏 = log 𝑥

6. Product Property of Logs
This is known as the Product Property
log 𝑏 𝑥 + log 𝑏 𝑦 = log 𝑏 𝑥𝑦
Any log of a number has the same value as the sum of its factors.
In order for this to work they must have the same base.

7. Practice
Rewrite the following Log expression in as many different ways you can think of:
log 36

8. Figuring out the Log Property
Use your calculator to solve for x below:
log 20 − log 5 = log 𝑥
log 30 − log 3 = log 𝑥
log 5 − log 2 = log 𝑥
log 17 − log 9 = log 𝑥
log − log 17 = log 𝑥
log 𝑎 − log 𝑏 = log 𝑥

9. Quotient Property of Logs
This is known as the Quotient Property
log 𝑏 𝑥 − log 𝑏 𝑦 = log 𝑏 𝑥 𝑦
In order for this to work they must have the same base.

10. Fill in the Blank (no calculator)
log 60 = log − ?
log = log − ?
? = log − log

11. Log Properties - Summary
There are three log properties that we learned about:
Power Property
Product Property
Quotient Property
Great Chart on p. 335 (could be noteworthy)

12. Proofs – Product and Power

13. First Back to Exponent Rules
Complete the two exponent rules below:
𝑥 𝑎 𝑥 𝑏 = ______
𝑥 𝑏 𝑥 𝑎 = ______
𝑥 (𝑎+𝑏)
𝑥 𝑏−𝑎

14. Applications of Properties
In chemistry, a solution’s pH is defined by the logarithmic equation  𝑝 𝑡 =− log (𝑡) , where t is the hydronium ion concentration in moles per liter. We usually round pH values to the nearest tenth.
Without using a calculator find the pH of a solution with a hydronium ion concentration of 4.5 x 10-5

15. Applications of Properties
Solution:
𝑝 4.5× 10 −5 =− log 4.5× 10 −5
=− log log 10 −5
=− log −5log 10
=− log −5 1
=− log −5
≈4.3

16. Applications of Properties
An initial number of bacteria presented in a culture is 10, This number doubles every hour.
1) Write a function to express the number of bacteria after t hours has passed.
2) How long will it take to get the bacteria number 100,000?

17. Applications of Properties
Solution:
1) 𝑏 𝑡 =10, 𝑡
2) 100,000=10, 𝑡
10= 2 𝑡
log 10 = log 𝑡
log 10 =𝑡 log 2
log 10 log 2 =𝑡

18. Applications of Properties; revisited
An initial number of bacteria presented in a culture is 10, This number doubles every 30 minutes.
1) Write a function to express the number of bacteria after t hours has passed.
2) How long will it take to get the bacteria number 100,000?

19. Applications of Properties
Solution:
1) 𝑏 𝑡 =10, 𝑡 30
2) 100,000=10, 𝑡 30
10= 2 𝑡 30
log 10 = log 2 𝑡 30
1= 𝑡 30 log 2
30=𝑡 log 2
30 log 2 =𝑡

20. For Tonight
Homework:
7-111, 114, 115, 118, 119, and 122