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Exponential and Logarithmic Functions By: Jeffrey Bivin Lake Zurich High School jeff.bivin@lz95.org Last Updated: January 2, 2006
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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
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y = 2 x Jeff Bivin -- LZHS xy = 2 x y = 3 x -21/41/9 ½1/3 011 123 249 3827 y = 3 x
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y = 2 x Jeff Bivin -- LZHS xy = 5 x y = 1 x -21/251 1/51 011 151 2251 31251 y = 3 x y = 5 x y = 1 x
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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
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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
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y = 2 x Jeff Bivin -- LZHS x y = (½) x y = (1/3) x -249 23 011 1½1/3 2¼1/9 31/81/27 y = 3 x y = 5 x y = 1 x y = (1/3) x y = (½) x
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Jeff Bivin -- LZHS
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f(x) = 2 x Jeff Bivin -- LZHS
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f(x) = 2 x-3 Jeff Bivin -- LZHS
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f(x) = 2 x+2 - 3 Jeff Bivin -- LZHS
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f(x) = -(2) x-4 – 2 Jeff Bivin -- LZHS
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f(x) = 2 -x = (1/2) x Jeff Bivin -- LZHS
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f(x) = (½) x-3 - 2 = (2) -x+3 - 2 Jeff Bivin -- LZHS
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y = 2 x Jeff Bivin -- LZHS xy -21/4 ½ 01 12 24 38 xy 1/4-2 ½ 10 21 42 83 x = 2 y
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How do we solve this exponential equation for the variable y? y = 2 x x = 2 y Jeff Bivin -- LZHS
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LOGARITHMS exponentiallogarithmic b > 0 A > 0 Jeff Bivin -- LZHS
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exponentiallogarithmic Jeff Bivin -- LZHS
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Evaluate Jeff Bivin -- LZHS
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Evaluate Jeff Bivin -- LZHS
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Evaluate Jeff Bivin -- LZHS
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Evaluate Jeff Bivin -- LZHS
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Evaluate Jeff Bivin -- LZHS
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Evaluate Jeff Bivin -- LZHS
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y = 2 x Jeff Bivin -- LZHS xy -21/4 ½ 01 12 24 38 xy 1/4-2 ½ 10 21 42 83 x = 2 y y=log 2 x
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Jeff Bivin -- LZHS x y = log 2 x 1/4-2 ½ 10 21 42 83 y = log 2 x y = log 3 x y = log 5 x x = 2 y
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Jeff Bivin -- LZHS x y = log ½ x 1/42 ½1 10 2 4-2 8-3 y = log ½ x x = (½) y
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Solve for x log 2 (x+5) = 4 2 4 = x + 5 16 = x + 5 11 = x Jeff Bivin -- LZHS
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Solve for x log x (32) = 5 x 5 = 32 x 5 = 2 5 x = 2 Jeff Bivin -- LZHS
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Evaluate log 3 (25) = u 3 u = 25 3 u = 5 2 ?????? Jeff Bivin -- LZHS
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Change of Base Formula Jeff Bivin -- LZHS
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Evaluate log 3 (25) = 2.930 Jeff Bivin -- LZHS
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Evaluate log 5 (568) = 3.941 Jeff Bivin -- LZHS
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Properties of Logarithms Product Property Quotient Property Power Property Property of Equality Jeff Bivin -- LZHS
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Product Property multiplicationaddition multiplicationaddition Jeff Bivin -- LZHS
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Product Property Jeff Bivin -- LZHS
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Quotient Property divisionsubtraction divisionsubtraction Jeff Bivin -- LZHS
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Quotient Property Jeff Bivin -- LZHS
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Power Property log b (m p ) log b (m p ) = plog b (m) p Jeff Bivin -- LZHS
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Power Property Jeff Bivin -- LZHS
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Property of Equality Jeff Bivin -- LZHS
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Expand product property power property Jeff Bivin -- LZHS
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Expand quotient property product property power property Jeff Bivin -- LZHS
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Expand quotient property product property power property distributive property Jeff Bivin -- LZHS
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Condense power property product property quotient property Jeff Bivin -- LZHS
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Condense group / factor product property quotient property Power property Jeff Bivin -- LZHS
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Condense re-organize group product property Power property quotient property Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x checks! Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x checks! Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x checks! fails The argument must be positive Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x checks! Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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Solve for x Jeff Bivin -- LZHS
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