Exponential and Logarithmic Functions By: Jeffrey Bivin Lake Zurich High School Last Updated: January 2, 2006.

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Presentation transcript:

Exponential and Logarithmic Functions By: Jeffrey Bivin Lake Zurich High School Last Updated: January 2, 2006

With your Graphing Calculator graph each of the following y = 2 x y = 3 x y = 5 x y = 1 x Determine what is happening when the base is changing in each of these graphs. Jeff Bivin -- LZHS

y = 2 x Jeff Bivin -- LZHS xy = 2 x y = 3 x -21/41/9 ½1/ y = 3 x

y = 2 x Jeff Bivin -- LZHS xy = 5 x y = 1 x -21/251 1/ y = 3 x y = 5 x y = 1 x

y = 2 x Jeff Bivin -- LZHS y = 3 x y = 5 x y = 1 x Determine where each of the following would lie? y=10 x y=4 x y = (3/2) x y = 10 x y = 4 x y = (3/2) x

With your Graphing Calculator graph each of the following y = 1 x y = (1/2) x y = (1/3) x Determine what is happening when the base is changing in each of these graphs. Jeff Bivin -- LZHS

y = 2 x Jeff Bivin -- LZHS x y = (½) x y = (1/3) x ½1/3 2¼1/9 31/81/27 y = 3 x y = 5 x y = 1 x y = (1/3) x y = (½) x

Jeff Bivin -- LZHS

f(x) = 2 x Jeff Bivin -- LZHS

f(x) = 2 x-3 Jeff Bivin -- LZHS

f(x) = 2 x Jeff Bivin -- LZHS

f(x) = -(2) x-4 – 2 Jeff Bivin -- LZHS

f(x) = 2 -x = (1/2) x Jeff Bivin -- LZHS

f(x) = (½) x = (2) -x Jeff Bivin -- LZHS

y = 2 x Jeff Bivin -- LZHS xy -21/4 ½ xy 1/4-2 ½ x = 2 y

How do we solve this exponential equation for the variable y? y = 2 x x = 2 y Jeff Bivin -- LZHS

LOGARITHMS exponentiallogarithmic b > 0 A > 0 Jeff Bivin -- LZHS

exponentiallogarithmic Jeff Bivin -- LZHS

Evaluate Jeff Bivin -- LZHS

Evaluate Jeff Bivin -- LZHS

Evaluate Jeff Bivin -- LZHS

Evaluate Jeff Bivin -- LZHS

Evaluate Jeff Bivin -- LZHS

Evaluate Jeff Bivin -- LZHS

y = 2 x Jeff Bivin -- LZHS xy -21/4 ½ xy 1/4-2 ½ x = 2 y y=log 2 x

Jeff Bivin -- LZHS x y = log 2 x 1/4-2 ½ y = log 2 x y = log 3 x y = log 5 x x = 2 y

Jeff Bivin -- LZHS x y = log ½ x 1/42 ½ y = log ½ x x = (½) y

Solve for x log 2 (x+5) = = x = x = x Jeff Bivin -- LZHS

Solve for x log x (32) = 5 x 5 = 32 x 5 = 2 5 x = 2 Jeff Bivin -- LZHS

Evaluate log 3 (25) = u 3 u = 25 3 u = 5 2 ?????? Jeff Bivin -- LZHS

Change of Base Formula Jeff Bivin -- LZHS

Evaluate log 3 (25) = Jeff Bivin -- LZHS

Evaluate log 5 (568) = Jeff Bivin -- LZHS

Properties of Logarithms Product Property Quotient Property Power Property Property of Equality Jeff Bivin -- LZHS

Product Property multiplicationaddition multiplicationaddition Jeff Bivin -- LZHS

Product Property Jeff Bivin -- LZHS

Quotient Property divisionsubtraction divisionsubtraction Jeff Bivin -- LZHS

Quotient Property Jeff Bivin -- LZHS

Power Property log b (m p ) log b (m p ) = plog b (m) p Jeff Bivin -- LZHS

Power Property Jeff Bivin -- LZHS

Property of Equality Jeff Bivin -- LZHS

Expand product property power property Jeff Bivin -- LZHS

Expand quotient property product property power property Jeff Bivin -- LZHS

Expand quotient property product property power property distributive property Jeff Bivin -- LZHS

Condense power property product property quotient property Jeff Bivin -- LZHS

Condense group / factor product property quotient property Power property Jeff Bivin -- LZHS

Condense re-organize group product property Power property quotient property Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x checks! Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x checks! Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x checks! fails The argument must be positive Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x checks! Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS

Solve for x Jeff Bivin -- LZHS