October 21, 2004
Lecturer: Dr Martin Kurth
Michaelmas Term 2004
Problem Sheet 2 – Solutions
1. All we need here is the formula
ax = (eln a )x = ex ln a ,
and some of the calculation rules for exponentials.
(a) 10x = ex ln 10
(b) 102x = e2x ln 10
(c) 102+x = 102 10x = 100ex ln 10
(d) 4x b2x = (22 )x b2x = 22x b2x = (2b)2x = e2x ln(2b)
(e) 108x log 5 = (10log 5 )8x = 58x = e8x ln 5
2. Here we have to proceed in two steps:
Step 1: Solve for x.
Step 2: Interchange x and y.
(a) Step 1: y = e2x ⇔ 2x = ln y ⇔ x = (1/2) ln y
Step 2: y = (1/2) ln x
(b) Step 1: y = 2ex ⇔ ex = y/2 ⇔ x = ln(y/2)
Step 2: y = ln(x/2)
(c) Step 1: y = ln(2x) ⇔ 2x = ey ⇔ x = (1/2)ey
Step 2: y = (1/2)ex
(d) Step 1: y = log(2x) ⇔ 2x = 10y ⇔ x = (1/2)10y
Step 2: y = (1/2)10x
(e) Step 1:y = cos(2x) ⇔ 2x = cos−1 (y) ⇔ x = (1/2) cos−1 (y)
Step 2: y = (1/2) cos−1 (x)
3. To define sine and cosine of θ, we draw a little illustrative picture:
r
¡
¡
¡
¡
¡
¡
θ
x
1
¡
¡
¡
¡
y
The sine and cosine functions are defined as
sin(θ) =
y
,
r
cos(θ) =
x
.
r
This gives us
sin2 (θ) + cos2 (θ) =
x2
1
y2
+
= 2 (x2 + y 2 ).
2
2
r
r
r
From Pythagoras’ theorem we know that
x2 + y 2 = r2 ,
and thus
sin2 (θ) + cos2 (θ) =
2
r2
= 1.
r2