5.6 solving exponential and logarithmic equations

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5.6 solving exponential and logarithmic equations

  1. 1. Solving Exponential and Logarithmic Equations
  2. 2. • By the definition of a function, if u = v and f is a function, then f(u) = f(v) [remember each input of a function has one and only one output!] • So, let’s say we have a function f(x) = b?, then if u = v then bu = bv for all real #s b>0 • Likewise, if we have a function f(x) = logb?, if u = v then logbu = logbv for all real #s b>0 • And because exponential and logarithmic functions are one-to-one (i.e. their inverses are also functions), then the converse is true: – If bu = bv then u = v – If logbu = logbv then u = v
  3. 3. Now, let’s use these statements, (and all the other properties for exponents and logarithms), to solve: Exponential and Logarithmic equations
  4. 4. Example 1: Powers of the Same Base Solve 8x = 2x+1 Can you create a common base? (23)x = 2x+1 23x = 2x+1 3x = x+1 2x = 1 x = ½ Graph Y=8x and Y=2x+1 to confirm
  5. 5. Example 2: Powers of Different Bases Solve 5x = 2 ln5x = ln2 xln5 = ln2 x = ln2/ln5 x = 0.6931/1.6094 x = 0.4307 Graph Y=5x and Y=2 to confirm
  6. 6. Example 3: Powers of Different Bases Solve 24x-1 = 31-x ln(24x-1) = ln(31-x) (4x-1)ln2 = (1-x)ln3 4xln2 – ln2 = ln3 – xln3 4xln2 + xln3 = ln3 +ln2 x(4ln2 + ln3) = ln3 +ln2 x = (ln3 + ln2)/(4ln2 + ln3) x = 0.4628 Graph Y=24x-1 and Y= 31-x to confirm
  7. 7. Example 4: Using Substitution Solve ex – e-x= 4 Multiply each side by ex to eliminate negative exponents ex(ex – e-x ) = ex(4) exex – exe-x = ex(4) e2x - 1 = 4ex e2x - 4ex - 1 = 0 Let u = ex and substitute u2 -4u -1 = 0
  8. 8. Now, let’s do the same thing by solving some real world applications of exponential equations
  9. 9. Example 5: Radiocarbon Dating The half life of carbon-14 is 5730 years. The skeleton of a mastodon has lost 58% of its original carbon-14. When did the mastodon die? What is the half-life formula? f(x) = P(0.5)x/h (P is initial amount; h is half-life, x is # years) So, f(x) = P(0.5)x/5730 .42P = P(0.5)x/5730
  10. 10. 0.42P = P(0.5)x/5730 0.42 = (0.5)x/5730 ln0.42 = ln(0.5)x/5730 ln0.42 = (x/5730)ln0.5 x = (5730)(ln0.42)/ln0.5 x = 7171.3171 Graph Y=0.42 and Y=(0.5)x/5730 to confirm
  11. 11. Example 6: Compound Interest If $3000 is to be invested at 8% per year, compounded quarterly, in how many years will the investment be worth $10,680? What is the compound interest formula? A = P(1+r)t (A is amount after t periods; P is the principal; r is interest rate) So, 10,680 = 3000(1+.08/4)4t 10,680 = 3000(1.02)4t
  12. 12. 10,680 = 3000(1.02)4t 3.56 = (1.02)4t ln3.56 = ln(1.02)4t ln3.56 = (4t)ln(1.02) 4t = ln3.56/ln1.02 t = (ln3.56/ln1.02)/4 t = 16.03 years Graph Y=10680 and Y=3000(1.02)t/4 to confirm
  13. 13. Example 7: Population Growth The population of a certain type of bacteria grows by the function S(t) = Pert, where P is the initial population and r is a continuous growth rate. If a biologist has a culture that contains 1000 bacteria, and 7 hours later there are 5000 bacteria: a. Write the function for this population b. When will the population reach 1 billion? a. What does the equation look like when evaluated at t = 7, i.e. when S(7) = 5000? 5000 = 1000er(7) 5 = e7r ln5 = lne7r ln5 = 7rlne (lne = 1) ln5 = 7r r = 0.2299 So the function for this population is S(t) = 1000e0.2299t
  14. 14. b. When will the population reach 1 billion? b. S(t) = 1000e0.2299t 1000e0.2299t = 1,000,000,000 e0.2299t = 1,000,000 lne0.2299t = ln1,000,000 0.2299t lne = ln1,000,000 (lne =1) t = ln1,000,000/0.2299 t = 60.0936 hours Graph Y=1000e0.2299t and Y=1 billion to confirm
  15. 15. Ex 8: Inhibited Population Growth A population of fish in a lake at time t months is given by: F(t)=20,000/(1+24e-t/4) How long will it take for the population to reach 15,000? 15000 = 20000/(1+24e-t/4) 15000(1+24e-t/4) = 20000 1+24e-t/4 = 20000/15000 24e-t/4 = 4/3 - 1 e-t/4 = (1/3)(1/24) lne-t/4 = ln (1/72) (-t/4)lne = ln1 - ln72 -t/4 = 0 – ln72 (lne=1, ln1=0) t = 4ln72 = 17.1067 Graph Y=15000 and Y=20000/(1+24e-t/4) to confirm
  16. 16. Now, let’s solve some logarithmic equations using the properties of one-to-one functions to help
  17. 17. Example 10: Equations with Logarithmic and Constant Terms Solve ln(x – 3) = 5 - ln(x – 3) First put all log terms on one side of equal sign & constants on the other ln(x – 3)+ ln(x – 3)= 5 2ln(x - 3) = 5 ln(x -3) = 5/2 e ln(x -3) = e5/2 x -3 = e5/2 x = e5/2 +3 = 15.1825 Graph Y= ln(x-3) and Y= 5 – ln(x-3) to confirm
  18. 18. Example 11: Equations with Logarithmic and Constant Terms Solve log(x – 16) = 2 - log(x – 1) log(x – 16)+ log(x – 1) = 2 log[(x-16)(x-1)] = 2 log(x2 -17x +16) = 2 10 log(x2 -17x +16) = 102 x2 -17x +16 = 100 x2 -17x – 84 = 0 (x +4)(x – 21) = 0 x = -4 and 21, but log(x-16) and log(x-1) are not defined for x = -4, so only x = 21 is valid Graph Y= log(x-16) and Y= 2 – log(x-1) to confirm

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