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0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
0410 ch 4 day 10
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0410 ch 4 day 10

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  • Transcript

    • 1. 4.4 Exponential & Logarithmic Equations Day 3Romans 15:13  May the God of hope fill you withall joy and peace in believing, so that by the powerof the Holy Spirit you may abound in hope.
    • 2. $2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly.
    • 3. $2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt ⎛ r ⎞ A = P ⎜ 1+ ⎟ ⎝ n ⎠
    • 4. $2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt ⎛ r ⎞ A = P ⎜ 1+ ⎟ ⎝ n ⎠ 12t ⎛ .035 ⎞4000 = 2000 ⎜ 1+ ⎟ ⎝ 12 ⎠
    • 5. $2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt ⎛ r ⎞ A = P ⎜ 1+ ⎟ ⎝ n ⎠ 12t ⎛ .035 ⎞4000 = 2000 ⎜ 1+ ⎟ ⎝ 12 ⎠ ⎛ .035 ⎞log 2 = 12t log ⎜ 1+ ⎟ ⎝ 12 ⎠
    • 6. $2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt ⎛ r ⎞ A = P ⎜ 1+ ⎟ ⎝ n ⎠ log 2 =t ⎛ .035 ⎞ 12t 12 log ⎜ 1+ ⎟ ⎛ .035 ⎞ ⎝ 12 ⎠4000 = 2000 ⎜ 1+ ⎟ ⎝ 12 ⎠ ⎛ .035 ⎞log 2 = 12t log ⎜ 1+ ⎟ ⎝ 12 ⎠
    • 7. $2000 is invested at an interest rate of 3.5% per year.Find the time for the money to double if the interest iscompounded monthly. nt ⎛ r ⎞ A = P ⎜ 1+ ⎟ ⎝ n ⎠ log 2 =t ⎛ .035 ⎞ 12t 12 log ⎜ 1+ ⎟ ⎛ .035 ⎞ ⎝ 12 ⎠4000 = 2000 ⎜ 1+ ⎟ ⎝ 12 ⎠ t ≈ 19.8 years ⎛ .035 ⎞log 2 = 12t log ⎜ 1+ ⎟ ⎝ 12 ⎠
    • 8. A sum of money invested at a fixed rate with continuouscompounding tripled in 19 years. Find the rate.
    • 9. A sum of money invested at a fixed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe
    • 10. A sum of money invested at a fixed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e
    • 11. A sum of money invested at a fixed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e ln 3 = 19r ln e
    • 12. A sum of money invested at a fixed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e ln 3 = 19r ln e ln 3 =r 19
    • 13. A sum of money invested at a fixed rate with continuouscompounding tripled in 19 years. Find the rate. rt A = Pe 19r 3 = 1e ln 3 = 19r ln e ln 3 =r 19 r ≈ 5.8%
    • 14. If I o and I denote the intensity of light before andafter going through a material, and x is the distance infeet the light travels in the material, then −1 ⎛ I ⎞ ln ⎜ ⎟ = x k ⎝ I o ⎠where k is a constant depending on the type of material.
    • 15. If I o and I denote the intensity of light before andafter going through a material, and x is the distance infeet the light travels in the material, then −1 ⎛ I ⎞ ln ⎜ ⎟ = x k ⎝ I o ⎠where k is a constant depending on the type of material. This is the Beer-Lambert Law
    • 16. If I o and I denote the intensity of light before andafter going through a material, and x is the distance infeet the light travels in the material, then −1 ⎛ I ⎞ ln ⎜ ⎟ = x k ⎝ I o ⎠where k is a constant depending on the type of material. This is the Beer-Lambert Law Solve the equation for I
    • 17. −1 ⎛ I ⎞ ln ⎜ ⎟ = xk ⎝ I o ⎠
    • 18. −1 ⎛ I ⎞ ln ⎜ ⎟ = xk ⎝ I o ⎠ ⎛ I ⎞ln ⎜ ⎟ = −kx ⎝ I o ⎠
    • 19. −1 ⎛ I ⎞ ln ⎜ ⎟ = xk ⎝ I o ⎠ ⎛ I ⎞ln ⎜ ⎟ = −kx ⎝ I o ⎠ − kx I e = Io
    • 20. −1 ⎛ I ⎞ ln ⎜ ⎟ = xk ⎝ I o ⎠ ⎛ I ⎞ln ⎜ ⎟ = −kx ⎝ I o ⎠ − kx I e = Io − kx I oe =I
    • 21. −1 ⎛ I ⎞ ln ⎜ ⎟ = xk ⎝ I o ⎠ ⎛ I ⎞ln ⎜ ⎟ = −kx ⎝ I o ⎠ − kx I e = Io − kx I oe =I or − kx I = I oe
    • 22. −1 ⎛ I ⎞ ln ⎜ ⎟ = x This is just a variation ofk ⎝ I o ⎠ rt A = Ao e ⎛ I ⎞ln ⎜ ⎟ = −kx or ⎝ I o ⎠ rt − kx I y = Pe e = Io − kx I oe =I (This is part of Example 10 or on page 364) − kx I = I oe
    • 23. HW #9It is literally true that you can succeed best andquickest by helping others to succeed. Napoleon Hill

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