Lesson 1 Log Logic Develop Understanding

Ready

Graph each function over the domain ${- 4 \leq x \leq 4}$ .

1.

$y = 2^{x}$

2.

$y = 2 \cdot 2^{x}$

3.

$y = {(\frac{1}{2})}^{x}$

4.

$y = 2 {(\frac{1}{2})}^{x}$

5.

Compare problem 1 to problem 2. Multiplying by $2$ should generate a vertical stretch of the graph, but the graph looks like it has been translated vertically. How do you explain that?

6.

Compare problem 3 to problem 4. Is your explanation in problem 5 still valid for these two graphs? Explain.

Set

7.

Given that $f (x) = 3^{x}$ , $f^{- 1} (x) =$

8.

Given that $f (x) = 7^{x}$ , $f^{- 1} (x) =$

9.

Given that $f (x) = a^{x}$ , $f^{- 1} (x) =$

10.

Given that $f^{- 1} (x) = \log_{10} x$ , $f (x) =$

11.

Given that $f^{- 1} (x) = \log_{27} x$ , $f (x) =$

Given $f (x) = 5^{x}$ . Use the table to fill in the missing values and evaluate the log expression.

$x$	$f (x) = 5^{x}$
$0$	$1$
$1$	$5$
$2$	$25$
$3$	$125$
$4$	$625$
$5$	$3,125$

12.

$f^{- 1} (25) = \log_{5}$ $=$ .

13.

$f^{- 1} (3,125) = \log_{5}$ $=$ .

14.

$f^{- 1} ($ $) = \log_{5}$ $= 1$

Given $h (x) = 2^{x}$ and $k (x) = h^{- 1} (x)$ .

Use the graph of $h (x) = 2^{x}$ to find the missing value in each equation.

15.

$h^{- 1} ($ $) = \log_{2} (\frac{1}{2}) =$ .

16.

$h^{- 1} (1) = \log_{2}$ $=$ .

17.

$h^{- 1} (0.25) = \log_{2}$ $=$ .

18.

Answer the question yes or no. If yes, give an example of the answer. If no, explain why not.

Does $\log_{x} 0$ have an answer?

Go

Apply the properties of exponents to find equivalent numerical expressions that no longer have exponents.

19.

$27^{0}$

20.

$11 {(- 6)}^{0}$

21.

$- 3^{- 2}$

22.

$4^{- 3}$

23.

$\frac{9}{2^{- 1}}$

24.

$\frac{4^{3}}{8^{0}}$

25.

$3 {(\frac{29^{3}}{11^{5}})}^{0}$

26.

$\frac{3}{6^{- 1}}$

27.

$\frac{32^{- 1}}{4^{- 1}}$