Review Exercises – Exponentials and Log’s(for practice as needed – not to hand in)1. Suppose ln w = u − ln 2.Express w in terms of u.2. Suppose ln |y| − ln |a − y| = kx + c.Express y as a function of x.3. a) y = y(t) = C e−ktwith k > 0, and for n = 2, 3, . . . let tnbe defined by theequation y(tn) =1ny(0). Find the formula for tn.b) Give a rough sketch of y = y(t) and the approximate location of the first few values oftn. What is the spacing of the values tnon the t-axis as n increases? How does thisspacing relate to the behavior of exponential growth and decay ?4. a) ComputeZ11 + e−ktdt b)Z1ekt+ e−ktdt.Hint: do something to the integrand first in each case. Laws of exponents are good too.(over for solutions)Exp & Log Exercises – Answers1. ln w = u − ln 2 → logew = u − loge2 → elogew= eu−loge2= eue− loge2→ w =eu2.2. ln |y| − ln |a − y| = kx + c → ln |ya−y| = kx + c → |ya−y| = ekx+c= C ekxwithC = ec> 0; to remove the absolute value, replace C > 0 by arbitrary C (with C 6= 0 sincey 6= 0 here. Thenya−y= C ekx+ algebra (and replacing C−1by C) givesy =a1 + Ce−kx.3. y = y(t) = C e−kta) y(tn) =1ny(0) → e−ktn=1n→ k tn= − ln n → tn=1kln n .b) The spacing between tnand tn+1gets smaller and smaller, and in fact goes to zero as ngets large (as can be seen using ln(n + 1) − ln n = ln(n+1n) = ln(1 +1n)). This reflects thefact that exponential decay is very fast; for example, if it takes tH= 1 time unit fory(1) =12y(0) (half-life), then y(10) = 2−10y(0), i.e. it takes only 10 time units for y todecay from y(0) to to approximately .001 y(0).4.Z11 + e−ktdt =Zekt1 + ektdt =1kln(1 + ekt) + c .Z1ekt+ e−ktdt =Zekt1 + e2ktdt =1ktan−1(1 + ekt) + c
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