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Page 1 of 10 Math 1330 Algebra Review Part 3 Exponential Functions Functions whose equations contain a variable in the exponent are called exponential functions. The exponential function with base is defined by ()= ( >0 β‰  1) and is any real number. If = (the natural base, e β‰ˆ 2.7183), then we have x e x f = ) ( . If > , the graph of ()= looks like (larger results in a steeper graph): If < < , the graph of ()= looks like (smaller results in a steeper graph): Domain:(βˆ’βˆž, ∞) Range:(0, ∞) 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. ()=3 +1 βˆ’ 4 b. ()=5 οΏ½ 1 2 οΏ½ +1

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Page 1: Math 1330 Algebra Review Part 3 Exponential Functions ...irina/MATH1330/1330Notes/1330...Algebra Review Part 3 . Exponential Functions . Functions whose equations contain a variable

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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 𝒇𝒇 with base 𝒃𝒃 is defined by 𝑓𝑓(π‘₯π‘₯) = 𝑏𝑏π‘₯π‘₯ (𝑏𝑏 > 0 π‘Žπ‘Žπ‘Žπ‘Žπ‘Žπ‘Ž 𝑏𝑏 β‰  1) and π‘₯π‘₯ is any real number. If 𝑏𝑏 = 𝑒𝑒 (the natural base, eβ‰ˆ2.7183), then we have xexf =)( . If 𝒃𝒃 > 𝟏𝟏, the graph of 𝑓𝑓(π‘₯π‘₯) = 𝑏𝑏π‘₯π‘₯ looks like (larger 𝑏𝑏 results in a steeper graph):

If 𝟎𝟎 < 𝒃𝒃 < 𝟏𝟏, the graph of 𝑓𝑓(π‘₯π‘₯) = 𝑏𝑏π‘₯π‘₯ looks like (smaller 𝑏𝑏 results in a steeper graph):

Domain:(βˆ’βˆž,∞) Range:(0,∞) 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. 𝑓𝑓(π‘₯π‘₯) = 3π‘₯π‘₯+1 βˆ’ 4

b. 𝑓𝑓(π‘₯π‘₯) = 5 οΏ½12οΏ½π‘₯π‘₯

+ 1

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c. 𝑓𝑓(π‘₯π‘₯) = βˆ’3(2π‘₯π‘₯+1) βˆ’ 5 Example: Write an equation of an exponential function with the following graph.

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Example: Given 𝑓𝑓(π‘₯π‘₯) = 2π‘₯π‘₯βˆ’1 and 𝑔𝑔(π‘₯π‘₯) = 5𝑒𝑒10π‘₯π‘₯. Evaluate (𝑔𝑔 ∘ 𝑓𝑓)(1) and (𝑔𝑔 ∘ 𝑓𝑓)(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 𝑓𝑓(π‘₯π‘₯) = log𝑏𝑏 π‘₯π‘₯ is the logarithmic function with base b with π‘₯π‘₯ > 0, 𝑏𝑏 > 0 and 𝑏𝑏 β‰  1. Note: The argument (inside) of a logarithmic function must be positive! Domain: (0,∞) Range: (βˆ’βˆž,∞) Key Point: (1,0) The Graph of a Logarithmic Function If 𝑏𝑏 > 1, the graph of 𝑓𝑓(π‘₯π‘₯) = log𝑏𝑏 π‘₯π‘₯ looks like:

If 0 < 𝑏𝑏 < 1, the graph of 𝑓𝑓(π‘₯π‘₯) = log𝑏𝑏 π‘₯π‘₯ looks like:

Both graphs have a vertical asymptote of x = 0 (the y-axis).

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Example: Evaluate π‘“π‘“βˆ’1(0) if 𝑓𝑓(π‘₯π‘₯) = 8βˆ’1βˆ’π‘₯π‘₯ βˆ’ 32 Example: Evaluate π‘“π‘“βˆ’1(βˆ’1) if 𝑓𝑓(π‘₯π‘₯) = 51βˆ’π‘₯π‘₯ βˆ’ 1 Logarithm Rules 𝑦𝑦 = log𝑏𝑏 π‘₯π‘₯ is equivalent to 𝑏𝑏𝑦𝑦 = π‘₯π‘₯. Example: log𝑏𝑏 1 = 0 For 𝑏𝑏 > 0, 𝑏𝑏 β‰  1, π‘₯π‘₯ > 0,𝑦𝑦 > 0: Inverse Property of Logarithms

1. log𝑏𝑏 𝑏𝑏π‘₯π‘₯ = π‘₯π‘₯

2. 𝑏𝑏log𝑏𝑏 π‘₯π‘₯ = π‘₯π‘₯ The Product Rule log𝑏𝑏(π‘₯π‘₯𝑦𝑦) = log𝑏𝑏 π‘₯π‘₯ + log𝑏𝑏 𝑦𝑦

The Quotient Rule log𝑏𝑏 οΏ½

π‘₯π‘₯𝑦𝑦� = log𝑏𝑏 π‘₯π‘₯ βˆ’ log𝑏𝑏 𝑦𝑦

The Power Rule log𝑏𝑏 π‘₯π‘₯𝑦𝑦 = 𝑦𝑦 log𝑏𝑏 π‘₯π‘₯

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Example: Evaluate, if possible. a. log7 49 b. log 1000 c. log2(βˆ’8)

d. log 0 e. ln π‘’π‘’βˆ’3 f. log7 √7

g. logπœ‹πœ‹ πœ‹πœ‹βˆ’βˆšπ‘’π‘’ h. log3 οΏ½19οΏ½

Example: Find the domain and vertical asymptotes of the following functions. a. ln(3 βˆ’ 2π‘₯π‘₯)

b. log3(π‘₯π‘₯2 βˆ’ π‘₯π‘₯ βˆ’ 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‘₯ βˆ’ 5)3

√π‘₯π‘₯ + 3οΏ½

Example: Rewrite the following expression as a single logarithm.

3 log7(π‘₯π‘₯ βˆ’ 2) βˆ’15

log7(π‘₯π‘₯2 βˆ’ 3) βˆ’ 4 log7(π‘₯π‘₯ + 5) + 1

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Example: Find (𝑓𝑓 ∘ 𝑔𝑔)(π‘₯π‘₯) when 𝑓𝑓(π‘₯π‘₯) = 𝑒𝑒3π‘₯π‘₯ and 𝑔𝑔(π‘₯π‘₯) = 5 ln π‘₯π‘₯. Example: Given 𝑓𝑓(π‘₯π‘₯) = 5𝑒𝑒4π‘₯π‘₯βˆ’1 and 𝑔𝑔(π‘₯π‘₯) = log2(4π‘₯π‘₯ + 1), find π‘“π‘“βˆ’1(5) + π‘”π‘”βˆ’1(2).

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Example: Solve for π‘₯π‘₯. 32

π‘₯π‘₯2 = 20

Example: Solve for π‘₯π‘₯.

2π‘₯π‘₯βˆ’1 =1

16

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Example: Solve for π‘₯π‘₯.

log(4π‘₯π‘₯ βˆ’ 1) + 3 = 5 Example: Solve for π‘₯π‘₯.

6𝑒𝑒π‘₯π‘₯ = 13

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Example: Find any π‘₯π‘₯- intercepts of the following function.

𝑓𝑓(π‘₯π‘₯) = log6(π‘₯π‘₯ + 5) + log6 π‘₯π‘₯ βˆ’ log5 25