math 1330 algebra review part 3 exponential functions ...irina/math1330/1330notes/1330...algebra...
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Math 1330 Algebra Review Part 3
Exponential Functions Functions whose equations contain a variable in the exponent are called exponential functions. The exponential function ππ with base ππ is defined by ππ(π₯π₯) = πππ₯π₯ (ππ > 0 ππππππ ππ β 1) and π₯π₯ is any real number. If ππ = ππ (the natural base, eβ2.7183), then we have xexf =)( . If ππ > ππ, the graph of ππ(π₯π₯) = πππ₯π₯ looks like (larger ππ results in a steeper graph):
If ππ < ππ < ππ, the graph of ππ(π₯π₯) = πππ₯π₯ looks like (smaller ππ results in a steeper graph):
Domain:(ββ,β) Range:(0,β) Key Point: (0,1) Both graphs have a horizontal asymptote of y = 0 (the x-axis). Example: Find the domain, range, and horizontal asymptote of the following functions.
a. ππ(π₯π₯) = 3π₯π₯+1 β 4
b. ππ(π₯π₯) = 5 οΏ½12οΏ½π₯π₯
+ 1
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c. ππ(π₯π₯) = β3(2π₯π₯+1) β 5 Example: Write an equation of an exponential function with the following graph.
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Example: Given ππ(π₯π₯) = 2π₯π₯β1 and ππ(π₯π₯) = 5ππ10π₯π₯. Evaluate (ππ β ππ)(1) and (ππ β ππ)(0). Logarithmic Functions Exponential functions are one-to-one; therefore, they are invertible. The inverse function of the exponential function with base b is called the logarithmic function with base b. The function ππ(π₯π₯) = logππ π₯π₯ is the logarithmic function with base b with π₯π₯ > 0, ππ > 0 and ππ β 1. Note: The argument (inside) of a logarithmic function must be positive! Domain: (0,β) Range: (ββ,β) Key Point: (1,0) The Graph of a Logarithmic Function If ππ > 1, the graph of ππ(π₯π₯) = logππ π₯π₯ looks like:
If 0 < ππ < 1, the graph of ππ(π₯π₯) = logππ π₯π₯ looks like:
Both graphs have a vertical asymptote of x = 0 (the y-axis).
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Example: Evaluate ππβ1(0) if ππ(π₯π₯) = 8β1βπ₯π₯ β 32 Example: Evaluate ππβ1(β1) if ππ(π₯π₯) = 51βπ₯π₯ β 1 Logarithm Rules π¦π¦ = logππ π₯π₯ is equivalent to πππ¦π¦ = π₯π₯. Example: logππ 1 = 0 For ππ > 0, ππ β 1, π₯π₯ > 0,π¦π¦ > 0: Inverse Property of Logarithms
1. logππ πππ₯π₯ = π₯π₯
2. ππlogππ π₯π₯ = π₯π₯ The Product Rule logππ(π₯π₯π¦π¦) = logππ π₯π₯ + logππ π¦π¦
The Quotient Rule logππ οΏ½
π₯π₯π¦π¦οΏ½ = logππ π₯π₯ β logππ π¦π¦
The Power Rule logππ π₯π₯π¦π¦ = π¦π¦ logππ π₯π₯
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Example: Evaluate, if possible. a. log7 49 b. log 1000 c. log2(β8)
d. log 0 e. ln ππβ3 f. log7 β7
g. logππ ππββππ h. log3 οΏ½19οΏ½
Example: Find the domain and vertical asymptotes of the following functions. a. ln(3 β 2π₯π₯)
b. log3(π₯π₯2 β π₯π₯ β 6)
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Example: Write the following logarithm as a sum of logarithms with no products, powers or quotients.
ln οΏ½π₯π₯4(π₯π₯ β 5)3
βπ₯π₯ + 3οΏ½
Example: Rewrite the following expression as a single logarithm.
3 log7(π₯π₯ β 2) β15
log7(π₯π₯2 β 3) β 4 log7(π₯π₯ + 5) + 1
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Example: Find (ππ β ππ)(π₯π₯) when ππ(π₯π₯) = ππ3π₯π₯ and ππ(π₯π₯) = 5 ln π₯π₯. Example: Given ππ(π₯π₯) = 5ππ4π₯π₯β1 and ππ(π₯π₯) = log2(4π₯π₯ + 1), find ππβ1(5) + ππβ1(2).
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Example: Solve for π₯π₯. 32
π₯π₯2 = 20
Example: Solve for π₯π₯.
2π₯π₯β1 =1
16
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Example: Solve for π₯π₯.
log(4π₯π₯ β 1) + 3 = 5 Example: Solve for π₯π₯.
6πππ₯π₯ = 13
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Example: Find any π₯π₯- intercepts of the following function.
ππ(π₯π₯) = log6(π₯π₯ + 5) + log6 π₯π₯ β log5 25