Precalculus: 4.4 More Equations and Applications

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1 Concepts: Solving Equations with Eponentials and Logarithms, eponential decay models with half-life radioactive dating, metabolization of drugs, compound interest formula, Newton s law of cooling. You have to be aware of etraneous solutions entering the problem when you are solving equations using eponentials and logarithms. In these cases, the etraneous solution can enter by finding a solution which is not in the domain of the original logarithmic function. Notation: e = ep. Technique: Try to isolate a single logarithm or eponential of and then take a logarithm or eponential to simplify. lnsome complicated function of = constant some complicated function of = e constant or e some complicated function of = constant some complicated function of = lnconstant In both cases, you solve for using what we have already learned about solving equations. Eample Solve the equation ln + ln = algebraically. ln + ln = ln = use ln A + ln B = lnab e ln = e take eponential of both sides of equation = e simplify using e ln A = A e = 0 quadratic in = b ± b 4ac a = ± 41 e 1 = ± 4 + 4e = ± e = ± 1 + e = 1 ± 1 + e However, e < 0, and in the original equation we had to have > for the logarithms to be defined. Therefore, the only solution is = e. Eample Solve the equation ln = for. ln = e ln = e take eponential of both sides = e simplify using inverse function rules, e ln A = A = ± e = ±e Page 1 of

2 Incorrect incomplete solution ln = ln = ln = 1 e ln = e 1 = e We missed the = e solution! This happened since the domain changed when we wrote ln = ln. Domain of ln is > 0 which means, 0 0,. Domain of ln is > 0 which means 0,. More properly, we should have written ln = ln since domain of ln is, 0 0,. ln = ln = ln = 1 e ln = e 1 = e = ±e Eample Solve the equation 44 = 3 algebraically e 7 Start by isolating the e 7 : 44 = e e 7 = e 7 = e 7 = e 7 = e 7 = 3 8 Now take logarithms of both sides: e 7 = Use ln e A = A: ln e 7 = ln = ln 3 You can have different forms of solution: = ln 3 = 1 7 ln 3 This is in the domain of the original logistic equation, so it is a solution. = 1 7 ln Page of

3 Eample Solve the equation ln ln = 3 algebraically. Start by writing as a single logarithm, use ln A ln B = lna/b: ln ln = 3 ln = 3 Now take eponential of both sides: e ln = e 3 Now solve for : = e 3 = e 3 e 3 = 1 e 3 = = 1 e 3 Since e >, this number is actually less than zero. But in our original equation, we had to have > 0 and > 0 > for the logarithms to be defined. These are both satisfied if >. So, sadly, this equation has no solution, since the only solution we found was not greater than. Eample Solve the equation ln = 3 algebraically. ln = 3 e ln = e 3 = e 3 = e 3 e 3 = 1 e 3 = = or, as above, we can figure out this is less than 0. 1 e3 Now, in our original equation, we require / > 0 for the logarithm to be defined. This is an inequality! We can solve it using a sign chart. The numerator is zero if =, the denominator is zero if = 0. These are the possible values where the function will change sign. positive 0 + negative positive From the sign chart, we see that the inequality is satisfied if, 0,. This is the domain of the function ln. Since our solution = is in the domain, it is not etraneous, it is a solution to the original problem. 1 e3 Page 3 of

4 Eample Solve log 3 + log 3 19 = log 3 0. First, notice the bases are all the same if they weren t, this would be mighty difficult to solve! Isolate a single logarithm: log 3 + log 3 19 = log 3 0 log 3 19 = log 3 0 log 3 19 log 3 0 = 0 19 log 3 = 0 0 Use 3 log 3 A = A: 3 log = = 1 0 = 39 Since = 39 is in the domain of the functions in the original equation, this is a solution. Eample Solve e e 1 = 0. This is a tricky one to solve, since it requires us to recognize that it is actually a quadratic in e! Here s the solution: Let z = e. Then z = e. Our equation becomes z z 1 = 0 z = b ± b 4ac a = 1 ± = 1 ± Now, we can get : Since z = e, we have = ln z = ln = ln 1 The solution with = ln is etraneous, since this is not a real number the number is negative, which is outside the domain of the logarithm. The only solution is = ln 1+. Eample How long does it take for 1g of carbon-14 in a tree trunk to be reduced to 10g of carbon-14 by radioactive decay? The half-life for carbon-14 is 730 years. The model for radioactive decay is At = A 0 e rt. First, use the half-life to determine r: At = A 0 e rt 1 A 0 = A 0 e r730 1 = er730 ln 1 = lner730 ln 1 = r730 r = ln = Page 4 of

5 The model becomes At = A 0 e t. Now put in the value for A 0 and At and solve for t: 10 = 1e t ln10/1 = lne t ln10/1 = t t = ln10/ = There will be 10g left after 107 years. Eample The level of a prescription drug built up in the human body over time can be found from the formula L = D 1 0. n/h where D is the amount taken every n hours and h is the drugs half-life in hours. If a doctor wants the level of the drug lorazepam half-life 14 hours to build up to a level of.8 milligrams in a patient taking. milligram doses, then how often should the doses be taken? D L = 1 0. n/h..8 = 1 0. n/ n/14 = n/14 = 1..8 = n/14 = ln[0. n/14 ] = ln[0.1971] n ln[0.] = n = = ln[0.] The patient should take the dosage every 1 hours. Page of