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The Exponential and Logarithmic Functions
Tada!
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3.1a The Exponential Model The Parent Function: π π₯ = π΅ π₯
The Transformations: π π₯ = ππ΅ π(π₯ββ) +π The VIPβs: Other stuff: π¦=0 (0,1) (1,π΅)
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3.1b The Exponential Model Why does it look like it does? π π₯ = 1 3 π₯
π π₯ = π₯ π π₯ = π₯ π π₯ = 2 π₯ x y -2 -1 1 2 3 x y -2 -1 1 2 3 x y -2 -1 1 2 3
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3.1b The Exponential Model Why does it look like it does?
-2 128 -1 32 8 1 2 1 2 3 1 8 x y -2 4 9 -1 4 3 4 1 12 2 36 3 108 π π₯ = π₯ π π₯ = 4 3 π₯
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3.1b Base as an obvious multiplier π(π₯)=π (πππ π) π₯ π=ππππ‘πππ π£πππ’π
(doubled, cut in half, etc.) π(π₯)=π (πππ π) π₯ π=ππππ‘πππ π£πππ’π (π¦βπππ‘ππππππ‘ π€βππ ππ β ππ π πππ£πππππ‘) πππ π=ππ’ππ‘ππππππ (πππ π)>1 β ππ₯ππππππ‘πππ ππππ€π‘β Assuming a is greater than 0 0<(πππ π)<1 β ππ₯ππππππ‘πππ πππππ¦ Assuming a is greater than 0
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3.1b The Natural Base π=2.71828182845904523 π= lim π₯ββ 1+ 1 π₯ π₯
Leonhard Euler Andrew Jackson π= π= lim π₯ββ π₯ π₯ βeβ¦isβ¦a little less than 3β π€ π π£ = π π£ π€ The Logistic Function π βπ = 1 π π π(π₯)= 1 1+ π βπ₯
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3.2a Base as an obvious multiplier Base as a percentage rate
(Population, radioactive decay) 3.2a Base as an obvious multiplier (doubled, cut in half, etc.) Base as a percentage rate π(π₯)=π (πππ π) π₯ π(π‘)= π 0 (1+π) π‘ π=ππππ‘πππ π£πππ’π π 0 =π=ππππ‘ππ π£πππ’π πππ π=ππ’ππ‘ππππππ π=πππππππ‘πππ πππ‘π Exβs 1-4 in book pg. 265 π‘=π‘πππ, π’π π’ππππ¦ π¦ππππ ππ’π‘ πππ¦ ππ‘π‘πππ‘πππ (πππ π)>1 β ππ₯ππππππ‘πππ ππππ€π‘β β(1+π)>1 0<(πππ π)<1 β ππ₯ππππππ‘πππ πππππ¦ β0<(1βπ)<1
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3.3a The Logarithmic Model What is it? A logarithm IS the exponent!
π₯π¨π π π=π is equivalent to π=π π log 2 8 = log = Some Properties log = log π 1=0 log π π=1 log π π π¦ =π¦ π log π π₯ =π₯ log = π€ π π£ = π π£ π€ log 4 1 = π βπ = 1 π π log 7 7 =
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3.3a The Common Log log 10 π₯ = log π₯ The Natural Log log π π₯ = ln π₯
Base is 10 The Natural Log log π π₯ = ln π₯ βeβ¦isβ¦a little less than 3β Base is e
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3.3b The Logarithmic Model The Parent Function: π π₯ = log π΅ π₯
The Transformations: π π₯ = alog π΅ (π π₯ββ ) +π The VIPβs: Other stuff: π₯=0 (1,0) (π΅,1)
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Properties of Logarithms
log π ππ = log π π + log π π log π π π = log π π β log π π log π π π =π log π π π€ π π£ = π π£ π€ log π 1=0 log π π=1 log π π π¦ =π¦ π log π π₯ =π₯ π βπ = 1 π π
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Solving Logarithms or for variables in the exponent
Make the bases the same (sometimes works) π π = π π π₯π¨π π π = π₯π¨π π π Rewrite (works nearly all of the time) π₯π¨π π π=π is equivalent to π=π π Take the log of both sides (always works)
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Graphing Logarithms/Exponents
Exponential Logarithmic π¦=π log π (π₯ββ) +π π(π₯)=π π π₯ββ +π π(π₯)=π π π₯ π(π‘)=π (1+π) π‘ π=ππππ‘πππ π£πππ’π π=% π=ππππ€π‘β πππ‘π π> ππ₯ππππππ‘πππ ππππ€π‘β βπ=(1+π) 0<π< ππ₯ππππππ‘πππ πππππ¦ βπ=(1βπ) Simplifying and Solving Make the bases the same (sometimes works) π π = π π π₯π¨π π π = π₯π¨π π π Rewrite (always works) π₯π¨π π π=π is equivalent to π=π π
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