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−kt with k > 0, and for n = 2, 3, . . . let tn be defined by the
equation y(tn ) = n1 y(0). Find the formula for tn .
b) Give a rough sketch of y = y(t) and the approximate location of the first few values of
tn . What is the spacing of the values tn on the t-axis as n increases? How does this
spacing relate to the behavior of exponential growth and decay ?
Z
1
1
dt
b)
dt.
1 + e−kt
ekt + e−kt
Hint: do something to the integrand first in each case. Laws of exponents are good too.
Z
4. a) Compute
(over for solutions)
Exp & Log Exercises – Answers
1. ln w = u − ln 2 → loge w = u − loge 2 → eloge w = eu−loge 2 = eu e− loge 2
eu
→ w =
.
2
y
y
| = kx + c → | a−y
| = ekx+c = C ekx with
2. ln |y| − ln |a − y| = kx + c → ln | a−y
C = ec > 0; to remove the absolute value, replace C > 0 by arbitrary C (with C 6= 0 since
y
y 6= 0 here. Then a−y
= C ekx + algebra (and replacing C −1 by C) gives
a
.
y =
1 + Ce−kx
3. y = y(t) = C e−kt
a) y(tn ) =
1
n
y(0) → e−ktn =
1
n
→ k tn = − ln n →
tn =
1
ln n .
k
b) The spacing between tn and tn+1 gets smaller and smaller, and in fact goes to zero as n
1
gets large (as can be seen using ln(n + 1) − ln n = ln( n+1
n ) = ln(1 + n )). This reflects the
fact that exponential decay is very fast; for example, if it takes tH = 1 time unit for
y(1) = 21 y(0) (half-life), then y(10) = 2−10 y(0), i.e. it takes only 10 time units for y to
decay from y(0) to to approximately .001 y(0).
Z
4.
Z
1
dt =
1 + e−kt
1
dt =
ekt + e−kt
Z
Z
ekt
dt =
1 + ekt
ekt
dt =
1 + e2kt
1
ln(1 + ekt ) + c .
k
1
tan−1 (1 + ekt ) + c .
k