|
|
MATH 1090 Algebra and Trigonometry
Problem Set 6 - Solutions
2002
Winter
- Sketch the graphs of
- y = 3x -
1
Graph is that of y = 3x
after a horizontal shift 1 unit to the right.
- y = 3x - 1
Graph is that of y = 3x
after a vertical shift 1 unit down.
Its horizontal asymptote is y =
- 1
- y = 1 - 3-x
Graph is that of y = 3x
after :
reflection in the y - axis,
reflection in the x - axis,
then vertical shift up by 1 unit
Horizontal asymptote: y = 1
- Evaluate, without using a calculator,
-
- log420 - log45
log420 - log45
=
- eln 5 + 2 ln 3
eln 5 + 2 ln 3 =
eln 5 + ln 32
=
eln (5 × 9) =
eln 45 = 45
OR
eln 5 + 2 ln 3 =
eln 5 × e2 ln 3
=
eln 5 × eln (3)2
= eln 5 × eln 9
=
5 × 9 = 45
Simplify for general x > -1 and evaluate for x =
1 :
-
x = 1
Þ
Therefore
- Write down the domain and range of
- f (x) = ln (x + 3)2
( x + 3 )2 > 0
x except
x = -3
Domain of f is
{ x | x
-3 }
Range of f is
- f (x) = 2 ln (x + 3)
f (x) = 2 ln (x + 3) :
x + 3 > 0 for x
> -3 only.
Domain of f is
{ x | x > -3 }
Range of f is
Note how the domains are different, even though
ln (x + 3)2
2 ln (x + 3) for
x > -3
- Express the solution set to these equations in terms of natural
logarithms (or e) and, where necessary, use a calculator to obtain a
decimal approximation, correct to three decimal places, for the solution.
- 10x = 17.65
Using common logarithms (base 10)
Þ x =
log 17.65 = 1.247
(3 d.p., by calculator)
In terms of natural logarithms:
10x = 17.65
Þ ln(10x)
= ln(17.65)
Þ
x ln(10) = ln(17.65)
Þ
= 1.247...
- e4x - 5 - 7 = 11 243
Þ e4x
- 5 = 11 243 + 7 = 11250
Þ 4x -
5 = ln(11250)
Þ 4x =
5 + ln(11250)
Þ (3 d.p.)
- 3x/7 = 0.2
Þ ln(3x/7)
= ln(0.2)
Þ
Þ
(3 d.p., by calculator)
Note that ln(0.2) = ln(5-
1) = -ln(5).
is an equivalent answer.
- e4x - 3
e2x - 18 = 0
Let t = e2x
then t2 - 3t - 18 = 0
Þ (t
- 6)(t + 3) = 0
Þ t =
e2x = 6   or
- 3
But e2x > 0, for all real x
e2x =
6 only
Þ 2x = ln(6)
Þ
(3 d.p., by calculator)
- log2(x - 1) +
log2(x + 1) = 3
Þ
log2((x - 1)(x + 1)) = 3
Þ
log2(x2 - 1)
= 3
Þ
x2 - 1 =
23 = 8
Þ
x2 = 9
Þ
x = ±3
BUT log2(-3
- 1)
is not real, so that
x = -3 is not in the domain of
log2(x - 1) +
log2(x + 1).
x = 3 is in the domain.
the only solution is
x = 3
Check: log2(3 - 1) +
log2(3 + 1)
= log2(2) + log2(4) = 1 + 2 = 3
- log2(4x + 1) = 5
Þ
4x + 1 = 25 = 32
Þ
4x = 31
Þ
(exactly)
-
Þ ln(x + 4) = 1
Þ
ln(x + 4) = 1 Þ
ln(x + 4) = 2
Þ x + 4 = e2
x =
e2
- 4 = 3.389
(3 d.p., by calculator)
- The number n(t) of radioactive atoms
remaining at time t is
where no = n(0) is the
initial number of radioactive atoms at time t = 0
and h is the half-life. In each
interval of one half-life, half of all remaining radioactive atoms decay.
- What proportion of the radioactive atoms remains after six half-lives?
of the atoms remain after 6 half-lives
(approximtely 1.6%)
- How many half-lives does it take before less than one-millionth of
the original number of radioactive atoms remains?
Note: y = ln x is an increasing function
of x everywhere on the domain.
The sense of the inequality is therefore preserved when one takes the
logarithm of both sides.
it takes
NEARLY 20 HALF-LIVES before less
than one-millionth
of the original atoms remain.
- [Challenge question:]
The exponential decay law can also be expressed as
n(t) = no e-kt
Find k in terms of h.
One may cancel t because t > 0
OR
[Return to your previous page]
Created 2002 03 04 and modified 2002 05 09 by Dr. G.H. George,
with assistance from Daniel Mastropietro.