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

Properties of Logs Pt. 2 Section 7.2.2

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

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

Figuring out the Log Property Use your calculator to solve for x below: log 5 + log 6 = log 𝑥 log 5 + log 2 = log 𝑥 log 3 + log 0.5 = log 𝑥 log 11 + log 17 = log 𝑥 log 9 + log 𝜋 = log 𝑥 log 𝑎 + log 𝑏 = log 𝑥

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.

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

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 375 − log 17 = log 𝑥 log 𝑎 − log 𝑏 = log 𝑥

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.

Fill in the Blank (no calculator) log 60 = log 120 − ? log 5 16 = log 5 8 − ? ? = log 24 75 − log 25 1.5

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)

Proofs – Product and Power

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

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

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

Applications of Properties An initial number of bacteria presented in a culture is 10,000. 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?

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

Applications of Properties; revisited An initial number of bacteria presented in a culture is 10,000. 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?

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

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