AMB111F Notes 4
Powers, Logs and their Graphs
Let a > 0. The number ax is called a power of x; x is the exponent or index
of the power; while a is the base of the power. If x = n, a natural number,
then an = a × a × · · · × a, (n times). E.g. 34 = 3 × 3 × 3 × 3 = 81. If x = −n,
1
1
, (n times). E.g. 3−4 = 3×3×3×3
= 1/81.
then a−n = a×a×···×a
Multiplication of Powers of the same base:
a ×am = |a × a ×
· · × a} = aa+m . E.g. 23 ×22 = 23+2 = 25 = 32.
{z· · · × a} × a
| × ·{z
n
n f actors
m f actors
Division
an ÷ am = an−m . E.g. 35 ÷ 33 = 35−3 = 32 = 9.
NOTE: (1) We define a0 = 1 for
a > 0, a 6= 1. (2) a1 = a; (3) a(b + c) = ab√+ ac
√
1/n
n
(distributive law); (4) a = a and a−1/n = a11/n ; (5) am/n = (am )1/n = n am ;
(6) (am an )t = ant bnt ; (7) If am = an where a > 0 and a 6= 1, then m = n. E.g.
If 2x = 22x−3 , then x = 2x − 3, implying x = 3.
(8) If ax = bx for a, b > 0, a, b 6= 1 then a = b. (9) If xn = a, then
√
(xn )1/n = a1/n, thus x = n a.
45x−3 ×3×754−x
25−x ×15x+2
(32×5)x−3 ×3×(3×52 )4−x
=
(52 )−x ×(3×5)x+2
53 × 3−3 = 125
27
Example 1. Simplify
x−3
4−x
Soln. 4525−x×3×75
×15x+2
32x−6+1+4−x−x−2 =
4×3x+1 −6×3x−1
3x+1 −3x
x
1 −6×3−1 )
4×3x+1 −6×3x−1
= 3 (4×3
3x+1 −3x
3x (31 −1)
=
32x−6 ×5x−3 ×3×34−x ×58−2x
5−2x ×3x+2 ×5x+2
= 5x−3+8−2x−(−2x)−x−2 ×
2. Simplify
Soln.
=
12−2
2
=5
3. Solve for x if (1/2)x−5 = 64.
Soln. 2−x+5 = 26 implies −x + 5 = 6, hence x = −1.
4. Solve for x if x−2/3 = 16.
Soln. (x−2/3)−3/2 = (24 )−3/2 , implying x = 2−6 = 1/64.
5. Solve for x if 3x+2 − 3x−1 = 78
Soln. 3x (32 − 1/3) = 78 implies 3x =
78×3
26
1
= 3 × 3 = 32 . Thus x = 2.
6. Express
√
8 as a power of 4.
Soln. If 81/2 = 4x , then 23/2 = 22x , implying 2x = 3/2, hence x = 3/4.
7. Solve for x and y if 4x+y = 2y+4 and x = 2y.
Soln. 22x+2y = 2y+4 implies 2x + 2y = y + 4, thus 2(2y) + 2y = y + 4. So
5y = 4, hence y = 4/5. Therefore x = 2y = 2(4/5) = 8/5.
8. Solve for x if 2x − 3x1/2 − 2 = 0.
Soln. Let x1/2 = y, so 2y 2√− 3y − 2 = 0 implies (2y + 1)(y − 2) = 0. Therefore
y = −1/2 or y = 2. But x = y = −1/2 is impossible, hence the only soln is
√
x = 2 implying x = 4.
√
x x 2
9. Solve for x if aa5−xa = 1.
x a2/x
Soln. aa5−x
= 1 implies ax+2/x−5+x = a0. Thus 2x2 + 2 − 5x = 0, so
√
1/2
(2x − 1)(x − 2) = 0. x cannot be 1/2 since
a2 is undefined. Hence x = 2.
10. Solve for x if 4 × 37x = 9 × 27x
Soln. (3/2)7x = (3/2)2 , thus 7x = 2 implying x = 2/7.
Logarithms
Recall: In ax , x is an exponent of ax. Now we also call x the logarithm of ax.
Here we always assume a > 0 and a 6= 1. We then write x = loga ax. (Read:
x is the log of ax with base a). Now let y = ax . Then x = loga y.
So we define y = loga x ⇔ x = ay . The first eqn is in logarithmic form while
the 2nd one is in exponential form. So we may convert from one form to the
other.
Examples: 1. (a) Evaluate log1/2 8; (b) if log10 x = 1, find x; (c) Evaluate :
log12 144 + log 2 32 − log 3 81 + log 1/3 27 ; (d) Express logb 2ax = t in exponential
form; (e) Express 26 = 64 in logarithmic form; (f) Evaluate log10 0.01.
Soln. (a) Let y = log1/2 8, then (1/2)y = 8 = 23 implying (2)−y = 23 . Thus
y = −3, i.e. log1/2 8 = −3.
(b) log10 x = 1 implies 101 = x, thus x = 10.
(c)Let y = log12 144, then 12y = 122 implying y = 2. Next let z = log2 32,
then 2z = 25 , implying z = 5. Next let a = log3 81, then 3a = 34 , thus a = 4.
2
Next let b = log1/3 27, then(1/3)b = 33 , i.e. 3−b = 33 , thus b = −3, therefore
y + z − a + b = 2 + 5 − 4 − 3 = 0.
(d) 2ax = bt
(e) log2 64 = 6
(f) Let y = log10 0.01, then 10y = 0.01 = 1/100 = 10−2 , thus y = −2.
Properties of Logs
Let a > 0 and a 6= 1. (i) loga ax = x
P roof. Let y = loga ax, then ay = ax , implying loga ax = y = x.
For example log3 84 == log3 34 = 4 (ii) loga 1 = 0
P roof. Let y = loga 1, then ay = 1 = a0, implying loga 1 = y = 0.
(iii) log1/a x = loga 1/x
P roof. Let y = log1/a x, then (1/a)y = x, implying a−y = x. Thus ay = 1/x.
Hence loga (1/x) = y, implying log1/a x = loga 1/x.
For example log1/2 8 = log2 (1/8) = log2 2−3 = −3.
(iv) loga a = 1.. P roof. Let y = loga a, then ay = a1 implies y = 1.
For example log7 7 = 1.
(v) loga ax = x loga a. P roof. By (i) loga ax = x = x loga a since loga a = 1 by
(iv).
(vi) loga x + loga y = loga (xy)
P roof. Let z1 = loga x and z2 = loga y, then az1 = x and az2 = y, implying
xy = az1 +z2 . Thus z1 + z2 = loga (xy) , i.e. loga (xy) = loga x + loga y.
(vii) Similarly loga x − loga y = loga (x/y)
(viii) loga bx = x loga b
P roof. Let y = x log a b, then y/x = loga b, implying ay/x = b. Thus ay = bx ,
therefore loga bx = y = x loga b
(ix) aloga y = y
P roof. Let z = loga y, then az = y, implying aloga y = y.
Example: 10log10 1/2 = 1/2.
(x) log x (without base) means log10 x. E.g. log 100 = log10 100 = log10 102 =
2.
3
More Examples
¯
16−log 9
1. Simplify log
.
log 4−log 3
log 16−log 9
= log(16/9)
log 4−log 3
log 4/3
(log 4−log 3)
log 16−log 9
= 2 log 4−log 3
log 4−log 3
−2 log3 7
Soln.
=
log(4/3)2
log 4/3
= 2/1 = 2
or
= 2.
2. Evaluate 3
−2
Soln. 3−2 log3 7 = 3log3 7 = 7−2 = 1/49.
3. If log 2 = 0.301 and log 3 = 0.477, find log 24; log 15 and log 0.81.
Soln. log 24 = log(3 × 23 ) = log 23 + log 3 = 3 log 2 + log 3 = 3(0.301) + 0.477 =
1, 38.
log 15 = log 5+log 3 = log(10/2)+log 3 = log10 10−log 2+log 3 = 1−(0.301)+
0.477 = 1, 176
log(0.81) = log(81/100) = log 81−log 100 = 4 log 3−2 = 4(0.477)−2 = −0.092
4. If log 72 = a and log 36 = b, find log 2 and log 3 in terms of a and b.
Soln. log 2 = log(72/36) = log 72 − log 36 = a − b
log 3 = log(36/12) = log 36 − log(3 × 22 ) = log 36 − log 3 − 2 log 2. Thus
log 2
= 3b−2a
.
log 3 = log 36−2
2
2
Change of Base
ca
If c > 0 and c 6= 1, then logb a = log
.
logc b
y
P roof. Let y = logb a, then b = a. Therefore logc by = logc a, thus y logc b =
ca
logc a, implying log
= y = logb a.
logc b
Example : 1. Use a calculator to find x if 3x = 2.
10 2
Soln. x = log3 2 = log
= 0.631 (to 3 decimal places).
log10 3
2. Find y such that log2 (y + 1) = logy+1 2.
22
Soln. Changing to base 2, we have log2 (y+1) = loglog(y+1)
, implying (log2 (y + 1))2 =
2
1. Therefore log2 (y + 1) = ±1, implying y + 1 = 2±1 . Thus y = −1 + 1/2 or
−1 + 2, i.e. y = −1/2 or y = 1.
3. Solve for x if log2 (x − 2) + log2 (x − 3) = 1.
Soln. log2 (x − 2)(x − 3) = 1, implying (x − 2)(x − 3) = 2. Thus x2 − 5x + 4 = 0,
i.e (x − 4)(x − 1) = 0. Since log2 (1 − 2) is undefined, we conclude that the
only correct solution is x = 4.
4
4. Solve for x if log x + log(x − 1) = log 12.
Soln. log x(x − 1) = log 12, implying x(x − 1) = 12. Thus x2 − x − 12 = 0,
i.e (x − 4)(x + 3) = 0. Since log(−3) is undefined, we conclude that the only
correct solution is x = 4.
5. Solve 32x√− 5 × 3x + 6 = 0.
Soln. 3x = 5± 25−4×6
= 3 or 2. Therefore x = 1 or x =
2
log 2
log 3
= 0.631.
Graphs of Exponential and Logarithmic Functions
Examples : 1. Graph y = 2x .
Soln. Consider the table of points on the graph:
x -3
-2
-1
0 1 2 3
y 1/8 1/4 1/2 1 2 4 8
An asymptote of a graph is a line such that the graph gets closer and closer to
it as x approaches a number or tends to infinity. An exponential graph with
eqn y = ax has an asymptote which is the negative x-axis if a > 1, otherwise
the asymptote is the positive x-axis.
In this example the asymptote is the negative x-axis. As x gets more negative,
the graph gets closer and closer to the x-axis. As x gets larger and larger, y
gets larger much faster than x, thus the graph bends towards the y-axis. So
the required graph is as shown in the sketch below:
y
•1
x
2. Graph y = (1/2)x .
Soln. Consider the table of points on the graph:
5
x -3 -2 -1 0 1
2
3
y 8 4 2 1 1/2 1/4 1/8
As discussed in the example above, the asymptote here is the positive x-axis.
As x gets more positive, the graph gets closer and closer to the x-axis. As x
gets more negative, y gets larger much faster than x, thus the graph bends
towards the y-axis. So the required graph is as shown in the sketch below:
y
1•
x
3. Graph y = log2 x.
Soln. In exponential form, the eqn is x = 2y . Then construct a table of some
points of the graph:
y -3
-2
-1
0 1 2 3
x 1/8 1/4 1/2 1 2 4 8
Notice that y is undefined at x = 0, so the line x = 0 (the y-axis) is an asymptote of the graph. As y gets more negative, the graph gets closer and closer
to the y-axis. As y gets more positive, x gets larger much faster than y, thus
the graph bends towards the x-axis. So the required graph is as shown in the
sketch below:
6
y
1
•
x
Note: The graphs of y = ax and y = loga x are symmetric about the line
y = x since they are mirror images of each other. Thus if a mirror is placed
along the line y = x, one sees behind the mirror the image of one graph which
is exactly the same as the other graph.
y = ax
y
y=x
•1
1
•
y = loga x
x
4. Graph y = log1/2 x.
Soln. In exponential form, the eqn is x = 2−y . Then construct a table of points
on the graph:
y -3 -2 -1 0 1
2
3
x 8 4 2 1 1/2 1/4 1/8
Notice that y = log2 (1/x) is undefined at x = 0, so the line x = 0 (the y-axis)
7
is an asymptote of the graph. As y gets more positive, the graph gets closer
and closer to the y-axis. As y gets more negative, x gets more negative much
faster than y, thus the graph bends towards the x-axis. So the required graph
is as shown in the sketch below:
y
1
•
x
5. Graph y = 21−x .
Soln. Consider the table of points on the graph:
x -3 -2 -1 0 1 2
3
y 16 8 4 2 1 1/2 1/4
The asymptote is the line y = 0. As x gets more positive, the graph gets
closer and closer to the x-axis. As x gets more and more negative, y gets larger
much faster than |x|, thus the graph bends towards the y-axis. So the required
graph is as shown in the sketch below:
8
y
2•
x
6. Graph y = −1 + 2x .
Soln. Consider the table of points on the graph:
x -3
-2
-1
0 1 2 3
y -7/8 -3/4 -1/2 -1 1 3 7
Here the asymptote is the line y = −1. As x gets more negative, the graph
gets closer and closer to the line y = −1. As x gets larger and larger, y gets
larger much faster than x, thus the graph bends towards the y-axis. So the
required graph is as shown in the sketch below:
y
•1
x
y = −1
7. Graph y = log2(1 − x).
Soln. In exponential form, the eqn is 1 − x = 2y , thus x = 1 − 2y . Then
construct a table of points on the graph:
9
y -3
-2
-1
0 1 2 3
x 7/8 3/4 1/2 0 -1 -3 -7
Notice that y is undefined at x = 1, so the line x = 1 is an asymptote of the
graph. As y gets more negative, the graph gets closer and closer to the line
x = 1. As y gets larger and larger, x gets larger much faster than y, thus
the graph bends towards the x-axis. So the required graph is as shown in the
sketch below:
y
0
•
1
x
1
8. Graph y = log2 (1−x)
.
Soln. In exponential form, the eqn is 1 − x = 2−y , thus x = 1 − 2−y . Then
construct a table of points on the graph:
y -3 -2 -1 0 1
2
3
x -7 -3 -1 0 1/2 3/4 7/8
Notice that y is undefined at x = 1, so the line x = 1 is an asymptote of the
graph. As y gets more negative, the graph gets closer and closer to the line
x = 1. As y gets larger and larger, x gets larger much faster than y, thus
the graph bends towards the x-axis. So the required graph is as shown in the
sketch below:
10
y
0
•
1
x
9. Graph y = 1 + log1/2(x − 1).
Soln. In exponential form, the eqn is x = 1 + 21−y . Then construct a table of
points on the graph:
y -3 -2 -1 0 1 2
3
x
9 5 3 2 3/2 5/4
Notice that y is undefined at x = 1, so the line x = 1 is an asymptote of
the graph. As y gets more positive, the graph gets closer and closer to the
line x = 1. As y gets more negative, x gets larger much faster than y, thus
the graph bends towards the x-axis. So the required graph is as shown in the
sketch below:
y
•
1
2
•
x
11
The Hyperbola
a
A graph of the eqn y = xna±b or y = b±x
n is called a hyperbola. It has asymptotes at xn ± b = 0 and at y = 0.
2
Examples : 1. Draw the graph of y = x−3
Soln. Asymptotes are the lines x = 3 and y = 0. Consider the following table
of points on the curve:
x -1
0
1 2 3
4 5 6
As x approaches 3, y apy -1/2 -2/3 -1 -2 undefined 2 1 2/3
proaches ±∞. As x gets larger, y approaches 0. Thus the graph is as shown
below:
y
3
x
2. Draw the graph of y = x22−2
√
√
Soln. Asymptotes are the lines x = 2, x = − 2 and y = 0. Consider the
following table of points on the curve:
√
x
-3
-2 -1 0 1 2 3
As x approaches ± 2, y approaches
y
2/7 1 -2 -1 -2 1 2/7
±∞. As x gets larger, y approaches 0. Thus the graph is as shown below:
12
y
−3−2−1
0 1 2 3
•−1
x
Log Inequalities
log is an increasing function, thus if loga x < loga y, then x < y whenever
a > 1. Examples : (1) Solve for x if log2 x < 3.
Soln. log2 x < 3 = 3 log 2 2 = log2 23 = log2 8. Thus the soln is 0 < x < 8 since
log is undefined for x ≤ 0.
(2) Solve for x if log1/2 x < 3.
Soln. log1/2 x = log2 (1/x) < 3 = 3 log 2 2 = log2 23 = log2 8. Thus 0 < 1/x < 8
since log is undefined for x ≤ 0. Therefore the soln is x > 1/8.
(3) Solve for x if log(x + 2) > log(1 − x).
Soln. Since a log is defined only for positive numbers, we have x + 2 > 0 and
1 − x > 0. We also have x + 2 > 1 − x since log is an increasing function. Thus
x > −2 and x < 1 and x > −1/2. Combining the three statements , we arrive
at the solution −1/2 < x < 1.
13