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3.2 – Logarithmic Functions and Their Graphs Ch. 3 – Exponential and Logarithmic Functions.

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Presentation on theme: "3.2 – Logarithmic Functions and Their Graphs Ch. 3 – Exponential and Logarithmic Functions."— Presentation transcript:

1 3.2 – Logarithmic Functions and Their Graphs Ch. 3 – Exponential and Logarithmic Functions

2 Logarithms Ex: 2 x = 512. Solve for x. How do you solve for an exponent? Use a logarithm! log 2 2 x = log 2 512 Don’t worry about how to solve that just yet, but know that logarithms are how we solve for an exponent x = 9! If a y = x, then log a x = y Scoop the loop!

3 Logarithmic Functions Logarithmic functions are always in the form: Ex: Graph f(x) = log 10 x by hand. ◦ log x = log 10 x ◦ Make a table and graph... ◦ Notable properties: ◦ Vertical asymptote at x=0 ◦ Domain: x > 0 ◦ Range: All real #s ◦ X – int: (1, 0) ◦ Note: f(x) = log a x is the inverse of f(x) = a x ! (a > 0, a ≠ 1) XY.25-.60.5-.30 10 2.30 4.60

4 Properties of Logs If no base is specified, assume the base is 10 e.g. log x = log 10 x log e x = ln x, which is called a natural logarithm log a 1 = 0 because a 0 = 1. log a a = 1 because a 1 = a. log a a x = x and a log a x = x because logs and exponents are inverse properties. If log a x = log a y, then x = y. log a x is only defined when x > 0.

5 Simplify: log 4 4 3 1. 1 2. 0 3. 3 4. 4 5. 16

6 Simplify: log 2 32 1. 30 2. 2 3. 16 4. 4 5. 5

7 Simplify: 1..01 2. -2 3. 3 4. 100 5. -3

8 Ex: Find the domain, asymptote(s), and intercept(s) of each function. To find domain of logs, set whatever is in the log to be greater than zero! The vertical asymptote will be the same as the value in the domain! Find the intercepts by setting one of the variables equal to zero, then solving! f(x) = log 3 (2x) Domain: 2x > 0, so x > 0 Asymptote: Since domain is x > 0, then asymptote is at x = 0 Intercepts: Y-int: none; can’t have zero in the log oAlso, y-axis is a vertical asymptote X-int: 0 = log 3 (2x)  3 0 = 2x  1 = 2x … o(½, 0)

9 Ex: Find the domain, asymptote(s), and intercept(s) of each function. f(x) = log(x + 2) + 1 Domain: x + 2 > 0, so x > -2 Asymptote: x = -2 Intercepts: Y-int: occurs when x = 0, so log(2) + 1  (0, 1.301) X-int: occurs when y = 0, so… o-1 = log(x+2)  10 -1 = x+2 .1 = x+2 … o(-1.9, 0)

10 Find the domain of f(x) = ln (x-2). 1. x > -2 2. x > 2 3. All real numbers 4. x ≥ 2 5. x ≥ -2 6. x ≠ -2 7. x ≠ 2

11 Find the x-int. of f(x) = ln (x-2) by hand. 1. (1, 0) 2. (2, 0) 3. (3, 0) 4. (e, 0) 5. None


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