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Some Solutions.......................................................................................... 1. This document is not complete. This is what is done NOW. MORE later. 2. The graph of y = ln(x – 2) + 3 is the same as that of y = lnx, but shifted 2 units to the right, so its domain is (2,4 ) rather than (0,4 ), then moved vertically 3 units (because y is 3 more than ln(x – 2) ). x=2 x=0 The graph of y = ln|x| is the same as the graph of y = ln x where x is positive, and where x is negative, it is just the reflection of y = ln x through the y-axis. y = ln |x – 1| is the graph of y = ln |x| , shifted 1 unit to the right, so the vertical asymptote occurs at x = 1, rather than at x = 0. 3. Simplify an expression using properties of logarithms. a. Simplify completely: 2log5(5x) + log5(5x) – 3log5 x 2log5(5x) + log5(5x) – 3log5 x log5(5x)2 + log5(5x) – log5 x3 (5x)2 (5x) log5 )))))))) x3 52 x2 5 x log5 )))))))) = log5 53 = 3 x3 b. Simplify: log2 (1/4) + log2 8 – log2 (3) + log2 (9) + log2 (.5) –2 + 3 + log2 ))) 9 ln 3 + –1 = log2 3 = )))))) 3 ln 2 4. Solve a logarithmic equation. log15 x + log15 (x – 2) = 1 log15 (x (x – 2)) = 1 x (x – 2) So x = 5 or –3 = 15 x2 – 2x – 15 = 0 But a quick inspection shows –3 is not a real solution, because log functions are not defined for negative input. (x – 5) (x + 3) = 0 5. Solve an exponential equation. a. Solve (answer must be exact): 2x+3 = 5x 2x+3 = 5x Exponential; haul out the logs ln( 2x+3) = ln (5x) and use those log properties! (x+3) ln 2 = x ln 5 x ln2 + 3ln2 = x ln 5 ln2 and ln5 are just numbers 3 ln2 = x (ln5 – ln2) 3 ln2 ln 5 – ln 2 ))))))))))) b. Solve: = x 2x = 3 ln 2x = ln3 x ln 2 = ln3 ln3 x = ))))) aka log ln 2 6. Solve a problem involving exponential growth. a. A bacterial population of 2400 has a relative growth rate of 4% per hour. How long will it be until the population reaches 14,400? Hmmm.... after 1 hour, there will be 2400(1+.04); after two hours, the number should be 2400(1+.04)C(1+.04) again After x hours, the number of bacteria will be 2400(1.04)x. So this question asks at what value of x will: 2400(1.04)x = 14400 (1.04)x = 14400/2400 = 6 ln (1.04)x = ln 6 x ln (1.04) = ln 6 x = ln 6 / ln 1.04 = 45.68 (hours) b. If you deposit $5000 into an account earning interest at the annual rate of 5%, compounded annually, how long will it take to triple? $5000 (1 + .05)t = $15000 (1 + .05)t = ln(1.05)t = ln 3 t ln(1.05) = ln 3 t c. 3 = ln 3/ ln1.05 Ñ 22.52 (years) How does answer to (b) change if the interest is compounded... daily? ...Continuously? After t years the value is $5000 (1 + .05/365)365t Daily: So, in a manner similar to above, we solve: (1 + .05/365)365t = 3 And we get 365tCln (1 + .05/365)365t = ln 3 ln3 t = ))))))))))))))) Ñ 21.97 (years) 365Cln (1 + .05/365) With continuous compounding, the computation is much neater: The balance after t years at annual rate “r” for initial deposit P is P e rt and we want that to be 3P. We solve e rt = 3 Taking ln of both sides: rt = ln 3 .... t = ln 4/r In this case r = .05, so t = ln 3/.05 (years) Ñ 21.97 yrs