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5-Minute Check on Activity 5-13

5-Minute Check on Activity 5-13. Use the properties of logarithms to simply the following expressions: ln xy log 10x/z ln x 5 log y7z3 ln (3xy5/z4) ln a + ln b log 5 – log 3. = ln x + ln y. = log 10 + log x – log z = 1 + log x – log z. = 5ln x. = 7log y – 3 log z.

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5-Minute Check on Activity 5-13

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  1. 5-Minute Check on Activity 5-13 Use the properties of logarithms to simply the following expressions: ln xy log 10x/z ln x5 log y7z3 ln (3xy5/z4) ln a + ln b log 5 – log 3 = ln x + ln y = log 10 + log x – log z = 1 + log x – log z = 5ln x = 7log y – 3 log z = ln 3 + ln x + 5 ln y – 4 ln z = ln ab = log 5/3 Click the mouse button or press the Space Bar to display the answers.

  2. Activity 5 - 14 Prison Growth

  3. Objectives • Solve exponential equations both graphically and algebraically

  4. Vocabulary • None new

  5. Activity Your sister is a criminal justice major at WCC. The following statistics appeared in one of her required readings relating to the inmate population of US federal prisons (population in thousands). She asks you to help analyze the prison growth situation for a project in her criminology course. Enter the data and plot using STATPLOT. What form does the graph look like? Exponential growth

  6. Activity cont Use an exponential regression model on your calculator to estimate the total inmate population equation. Graph the STATPLOT and the Y1 = on the same graph Use the model to estimate the prison population in 2010 P = abt = 11.36 (1.080)t P = 11.36 (1.080)t = 11.36(1.080)40 = 246.8 (thousand)

  7. Algebraic Method Solving exponential problems in form abx = c • Isolate exponential factor on one side of equationbx = c/a • Take the log (or ln) of both sides of equationln bx = ln (c/a) • Apply log property: log bx = x log b to remove variable as an exponentx ln b = ln c – ln a • Solve the resulting equation for the variablex = (ln c – ln a) / ln b

  8. Activity cont Solve 11.36(1.080)t = 180 algebraically 180 = 11.36 (1.080)t 15.8451 = (1.080)t ln 15.8451 = ln (1.080)t ln 15.8451 = t ln (1.080) ln 15.8451  ln (1.080) = 35.899 years = t 1970 + 35.899 ≈ 2006

  9. Activity cont Solve 11.36(1.080)t = 180 graphically y1 = 11.36 (1.080)t y2 = 180 Graph and find the intersection (2nd TRACE) x = 35.899 or about 2006

  10. Radioactive Decay Radioactive substances, such as uranium-235, stontium-90, iodine-131, and carbon-14, decay continuously with time. If P0 represents the original amount of a radioactive substance, then the amount P present after a time t (usually measured in years) is modeled by P = P0ekt where k represents the rate of continuous decay

  11. Radioactive Decay Example 1 One type of uranium decays at a rate of 0.35% per day. If 40 pounds of this uranium was found today, how much will be left after 90 days? P = P0ekt P = 40e-0.0035t P = 40e-0.0035(90) P = 40e-.315 = 29.2 lbs

  12. Radioactive Decay Example 2 Strontium-90 decays at a rate of 2.4% per year. If 10 grams of it were initially present, how much will be left after 20 years? P = P0ekt P = 10e-0.024t P = 10e-0.024(20) P = 10e-.48 = 6.2 grams

  13. Radioactive Decay Example 3 Strontium-90 decays at a rate of 2.4% per year. If 10 grams of it were initially present, how long until half of it is gone? (This is called half-life, a very important term with radioactive material) P = P0ekt For Note: (ln 0.5) ------------------ = half-life (decay rate) 5 = 10e-0.024t 0.5 = e-0.024t ln 0.5 = -0.024t (-0.69315)/(-0.024) = 28.88 years = t

  14. Summary and Homework • Summary • Solving exponential problems in form abx = c • Isolate exponential factor on one side of equation • Take the log (or ln) of both sides of equation • Apply log property: log bx = x log b to remove variable as an exponent • Solve the resulting equation for the variable • Graph as two functions and find the intersection • y1 = abx • y2 = c • Homework • pg 669 – 72; problems 1, 4-10

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