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Notes 6-4
 

Notes 6-4

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e and Natural Logarithms

e and Natural Logarithms

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    Notes 6-4 Notes 6-4 Presentation Transcript

    • Section 6-4 e and Natural Logarithms
    • Begin by opening your books to page 389. Pair up with a partner to work on the In-Class Activity. Make sure to record your observations on your note sheet.
    • e:
    • e: Comes out to be approximately 2.7182818284590...
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield e −1 r
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield e −1= e r .0735 −1
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63%
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63% b. The value of the investment after one year
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63% b. The value of the investment after one year 10,000 + 10,000(e .0735 − 1)
    • Example 1 If $10,000 is put into bonds that pay 7.35% interest compounded continuously, find: a. The annual yield e −1= e r .0735 − 1 ≈ .0762685367 So the annual yield is about 7.63% b. The value of the investment after one year 10,000 + 10,000(e .0735 − 1) ≈ $10,762.69
    • Continuous Change Model
    • Continuous Change Model When your principal P grows or decays continuously and an annual rate r, the amount A(t) after t years is:
    • Continuous Change Model When your principal P grows or decays continuously and an annual rate r, the amount A(t) after t years is: A(t) = Pe rt
    • Continuous Change Model When your principal P grows or decays continuously and an annual rate r, the amount A(t) after t years is: A(t) = Pe rt This means we have a model we can work with for continuous compounding, just like the other types of compounding we saw earlier in the year.
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period?
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe rt
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5)
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ $14,441.20
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ $14,441.20 b. How does this compare with the balance if the interest were compounded annually?
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ $14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) t
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ $14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) = 10,000(1.0735) t 5
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ $14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) = 10,000(1.0735) ≈ $14,256.41 t 5
    • Example 2 Suppose $10,000 is put into a 5-year certificate of deposit that pays 7.35% interest compounded continuously. a. What is the balance at the end of the period? A(t) = Pe = 10,000e rt .0735(5) ≈ $14,441.20 b. How does this compare with the balance if the interest were compounded annually? A = P(1+ r ) = 10,000(1.0735) ≈ $14,256.41 t 5 About $185 more
    • The Exponential Function with Base e:
    • The Exponential Function with Base e: A function of the form f(x) = ex
    • The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm:
    • The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm: The logarithm to the base e; written as ln x
    • The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm: The logarithm to the base e; written as ln x In other words, ln x = loge x
    • The Exponential Function with Base e: A function of the form f(x) = ex Natural Logarithm: The logarithm to the base e; written as ln x In other words, ln x = loge x This means that the Exponential Function with Base e and the Natural Logarithm are inverses of each other.
    • Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0?
    • Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0? ln7 − ln1
    • Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0? ln7 − ln1 ≈ 1.945910149
    • Example 3 Consider the region bounded by the following graphs: y = , the x-axis, the line x = a, the line x = b, x > 0. 1 x Using calculus, it can be proven that the area of that region is ln b - ln a. What then is the area bounded by the graphs: y= 1 x , the x-axis, the line x = 1, the line x = 7, x > 0? ln7 − ln1 ≈ 1.945910149 units 2
    • Homework
    • Homework p. 394 #1-20