# Week of Apr 16, 2018

Powerpoints: Apr 20 Apr 19 Apr 16

Monday
Topic: Direct and Inverse Variation
Classwork: Direct and Inverse Variation

Tuesday
Graphing (Mr. Hartzer absent)
Classwork: Graphing Polynomials and Rational Functions (Kuta), Extra Credit

Wednesday
Polynomials (Mr. Hartzer absent)
Classwork: Polynomial Review (Kuta), Extra Credit

Thursday
Topic: Rational Functions

Friday
Topic: Graphing Rational Functions
Classwork: Rational Functions Graphing 1

# Week of Apr 9, 2018

Powerpoints: Apr 13 Apr 12 Apr 9

Monday
Topic: Logarithm Review
Homework: Logarithm Review

Tuesday
SAT

Wednesday
SAT

Thursday
Topic: Unit 6 Quiz 2: Logarithms

Friday
Topic: Direct Variation Review
Classwork: Direct Variation Review

# Week of Apr 9, 2018

Powerpoints: Apr 12 Apr 9

Monday
Topic: Exponents and logarithms review

Tuesday
SAT

Wednesday
SAT

Thursday
Topic: Exponents and logarithms review
Classwork: Chapter 7 Honors Review

Friday
Topic: Unit 7 Test (Exponents and Logarithms)
Homework: Direct and Inverse Variation Review

# Week of Apr 2, 2018

NO SCHOOL: SPRING BREAK

# Week of Mar 26, 2018

Powerpoints: Mar 27

Monday
Topic: NO SCHOOL

Tuesday
Topic: Logarithm Properties
Classwork: 7-4 Properties of Logarithms

Wednesday
Topic: NO SCHOOL

Thursday
Topic: Test Review
Classwork: Unit 6 Study Guide

Friday
Topic: NO SCHOOL (Spring Break)

# Week of Mar 26, 2018

Powerpoints: Mar 27

Monday
Topic: NO SCHOOL

Tuesday
Topic: Logarithm Properties
Classwork: 7-4 Properties of Logarithms

Wednesday
Topic: Test Review
Classwork: Unit 6 Study Guide

Thursday
Topic: Unit 6 Test

Friday
Topic: NO SCHOOL (Spring Break)

# Unit 6 (Chapter 7) Test on Thursday

There will be a unit test on exponential functions and logarithms on Thursday, March 29. This test will not be moved; instead, it will be on whatever material we can cover by Tuesday. Retakes will have to be after the break.

# Week of Mar 19, 2018

Powerpoints: Mar 23 Mar 21 Mar 20 Mar 19

Monday
Topic: Quiz on Exponential Functions
Homework: Preparing for Logarithms

Tuesday
Topic: Logarithms

Wednesday
Topic: Logarithm Story Problems
Classwork: Logarithm Basics (Kuta)

Thursday
Topic: NO SCHOOL

Friday
Topic: Change of Base
Homework: Change of Base (Kuta)

# Week of Mar 19, 2018

Powerpoints: Mar 23 Mar 21 Mar 20 Mar 19

Monday
Topic: Quiz on Exponential Functions
Homework: Preparing for Logarithms

Tuesday
Topic: Logarithms

Wednesday
Topic: Logarithm Story Problems
Classwork: Logarithm Basics (Kuta)

Thursday
Topic: NO SCHOOL

Friday
Topic: Change of Base
Homework: Change of Base (Kuta)

# Logs and the Change of Base Formula

Prior to calculators, the standard way for determining a logarithm precisely was to use a look-up table for values. However, such tables were usually only printed for the common most bases, particularly 10 (log). To determine a logarithm for a different base, you needed to first convert to that base.

Luckily, it’s very simple to do so. The change of base formula is: $\log_a b = \frac{\log_c b}{\log_c a}$

This works for any base, but it’s most typical to use it for 10 and e. For instance, if you want to find $$\log_6 29$$ using a log (base 10) table, you find $$\log 6$$ and $$\log 29$$, then divide. That is, $\log_6 29 = \frac{\log 29}{\log 6} \approx \frac{1.4624}{.7782} \approx 1.879$

Side note: You’ll notice that the log table I linked only goes up to 9.99. So how did I use it to find $$\log 29$$? I looked up $$\log 2.9$$ and then added 1. This is why common log tables were most common for this purpose: They’re more flexible than natural log tables. You only need the values from 1.000 to 9.999 to find the log of any positive real number.

Since many calculators only have log and ln buttons, it’s useful to know this formula. Even for the calculator we use in class, you might find that it’s easier to use this technique instead of using [2nd][Window/F1].

You’ll get the same answer either way because the calculator is performing a change of base anyway. Using [2nd][Window/F1], find $$\log_0 0$$ and $$\log_1 1$$. You’ll get an error for each, but it’s a different error: In the first case, the error is Error: Domain because 0 is not a valid base. In the second case, the error is Error: Divide by 0. If the calculator were doing the logarithm directly, it should give another domain error (because 1 is not a valid base). By the change of base formula, though, you would calculate $$\log 1 \div \log 1 = 0 \div 0$$.

### Proof

The change of base formula seems weird: Why does it work? How does making the base into its own log work? This is magic!

It’s not magic, it’s math. Logarithms do sometimes work in ways we might not immediately predict. The proof is remarkably short, given how unusual it looks.

First, let’s replace each part of the formula with a variable. That is, we’ll let $x = \log_a b \\ y = \log_c b \\ z = \log_c a$

Let’s rewrite each of these into exponential form: $a^x = b \\ c^y = b \\ c^z = a$

Replacing $$a$$ in the first line with $$c^z$$ from the third line, then using the transitive property on the first two lines, gives us: $(c^z)^x = c^y$

Because of the properties of exponents, we can rewrite $$(c^z)^x = c^{xz}$$. Two exponential expressions with the same base (other than 1 and 0, which are excluded by definition) have to have the same exponent, that is: $xz = y \Rightarrow x = \frac{y}{z}$

We can now replace the variables as we defined them in the first step, giving us: $\log_a b = \frac{\log_c b}{\log_c a}$

which is the change of base formula.