Answers to activity questions
Topic: Definition of a Log
1.
Write the following in logarithmic form.
(a)
25 = 52
log 5 25 = 2
(b)
1024 = 45
log 4 1024 = 5
(c)
11.18 = 51.5
log 5 11.18 = 1.5
(d)
908.14 = 36.2
log 3 908.14 = 6.2
(e)
0.015625 = 8−2
log8 0.015625 = −2
(f)
490.5 = 7
0.4884 = 6−0.4
log 6 0.4884 = −0.4
(h)
0.5 = 4−0.5
log 4 0.5 = −0.5
(j)
1.568 = e0.45
ln1.568 = 0.45
(b)
log 2 1024 = 10
(g)
(i)
7.389 = e 2
log e 7.389 = 2 ( log e means ln)
7 = 490.5
log 49 7 = 0.5
ln 7.389 = 2
2.
(a)
Write the following in exponential form.
log100 = 2
log10 100 = 2
(c)
100 = 102
log 2.65 = 0.4232
2.65 = 10
(e)
(g)
log 4 100 = 3.322
100 = 43.322
(f)
0
log100 10 =
10 = 100
(d)
0.4232
ln1 = 0
1= e
1024 = 210
ln 20 = 2.996
20 = e 2.996
1
2
(h)
1
2
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ln 5 = 1.6094
5 = e1.6094
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3.
Evaluate the following logarithms using your calculator. Answer to 4 decimal places.
(a)
log 250
log 2.95
(b)
g250=2.3979
(c)
log 0.1
g2.95=0.4698
(d)
3
log  
8
g0.1= -1
g(3P8)= -0.4260
(e)
(g)
ln 3.05
h3.05= 1.1115
(f)
ln1000
h1000= 6.9078
ln 0.025
(h)
1
ln  
9
h0.025= -3.6889
h(1P9)= -2.1972
(i)
log 4 32
(j)
log 0.25 4
log 32
log 4
1.5051
=
0.6021
= 2.5
=
4.
=
g4Pg0.25= -1
Working in reverse, find the value of x.
(a)
log x = 0.7782
(c)
x = 100.7782
x=6
log x = −0.9031
log x = 2.8779
(b)
x = 102.8779
x = 754.9
(d)
log ( 2 x − 1) =
1.6902
−0.9031
x = 10
x = 0.125
(e)
log 4
log 0.25
ln x = 2.72
2x −1 =
101.6902
2 x − 1 =49 add 1 to both sides
2 x = 50 divide both sides by 2
x = 25
(f)
x = e 2.72
x = 15.18
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ln x = 0.75
x = e0.75
x = 2.117
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ln x = −0.5
(g)
(h)
3
4
ln 2 x = 0.75
ln 2 x =
−0.5
x=e
x = 0.6065
2 x = e0.75
2 x = 2.117
x = 1.0585
log 2 x = 2.72
(i)
(j)
log 5 x = −0.25
x = 22.72
x = 5−0.25
x = 6.589
x = 0.6687
Topic: The Logarithm Laws
1.
Evaluate the following without a calculator
(a)
(b)
log 2 32
log 5 125
= log 5 53
= log 2 25
= 3log 5 5
= 5log 2 2
=3
=5
(c)
1
log 2
16
= log 2 2−4
1
1
2−4
= =
16 24
(d)
log10 0.001
= log10 10−3
= −3log10 10
= −3
= −4
(e)
(f)
log 5 5
= log 5 5
log
1
2
= log n n
1
4
1
= log n n
4
1
=
4
2
2
2
=2
(h)
log n 4 n
2
( 2)
= 2 log ( 2 )
= log
1
= log 5 5
2
1
=
2
(g)
2
log x x
3
3
x
=
1
3 2
x )
(=
x
3
2
3
= log x x 2
3
log x x
2
3
=
2
=
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2.
(a)
Evaluate the following:
(b)
log 3 5 + log 3 1.8
= log 3 5 ×1.8
= log 5 (10 × 50 × 0.25)
= log 5 125
2
= log 3 9
(c)
log 5 10 + log 5 50 + log 5 0.25
=
log 3 9 log
=
2 log 3 3
33
=2
(d)
log 2 3 + log 2 24 − log 2 9
= log 2
= log 7
( 3 × 24 )
= log 2 8
9
24
24
= log 7 1
(f)
log 2 5 − log 2 10
5
10
= log 2
= log 2
32
36
1
= log 2
4
= log 2 2−2
= log 2
2× 5
1
2
= log 2 2
−
=0
2 log 2 3 − log 2 36
5
= log 2
( 8 × 3)
( 6 × 4)
= log 7
=3
(e)
=3
log 7 8 − log 7 6 − log 7 4 + log 7 3
1
2
= −2
1
2
5log 3 8 − 2 log 3 16
= −
(g)
(h)
85
162
215
= log 3 8
2
= log 3 27
3log10 5 − log10 2
= log 3
(i)
53
2
= log10 62.5
= log10
= 7 log 3 2
1
1
log 2 27 − log 2 36
3
2
1
3
= log 2 27 − log 2 36
(j)
1
2
= log 2 3 − log 2 6
3
6
= log 2 2−1
= log 2
= −1
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log 3 36
log 3 6
=
=
log 3 62
log 3 6
2 log 3 6
log 3 6
=2
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(k)
(l)
log 7 x
log 7 x x
1
2 log 5 8
3log 5 4
=
2 log 5 23
3log 5 22
=
6 log 5 2
6 log 5 2
=
log 2 3 × log 3 4
3
log 7 x 2
1
log 7 x
=2
3
log 7 x
2
1
1 2 1
=2= × =
3 2 3 3
2
=1
(m)
log 7 x 2
(n)
log 3 log 4
×
log 2 log 3
log 4
=
log 2
=
=
=
log 5 10 × log100 5
log 22
log 2
2 log 2
=
log10 log 5
×
log 5 log100
=
1
2
log 2
=2
3.
(a)
Write as a single Logarithm:
log x a + log x b − log x c
(b)
ab
= log x
c
2 log b x + log b
2 log a x − 2 log a y − 2 log a z
= 2(log a x − log a y − log a z )
= 2 log a
x
yz
1
2
2
= log b x + log b y + log b 7 − log b z 3
(d)
2
1
2
7x y
z3
1 + 2 log a b − log a ab
= log b
(c)
y + log b 7 − 3log b z
=log a a + 2 log a b − (log a a + log a b)
= log a a + 2 log a b − log a a − log a b
= log a b
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(e)
log 3 abc + log 3 b − 2 log 3 b
= log 3
(f)
2 log a ( x + 3) − log a ( x 2 − 9)
= log a ( x + 3) 2 − log a ( x 2 − 9)
a b2 c
b2
= log 3 ac
= log a
( x + 3) ( x + 3)
( x − 3) ( x + 3)
( x + 3)
( x − 3)
ln( x − 5) + 2
= ln( x − 5) + 2 ln e
= log a
(g)
2 log a a + 2
(h)
=4
= ln( x − 5) + ln e 2
= ln e 2 ( x − 5)
4.
(a)
(c)
Write in terms of individual logs
log x ab 2 c
(b)
=log x a + log x b 2 + log x c
 x2 y 
log a 

 z 
=log x a + 2 log x b + log x c
= log x x 2 + log x y − log x z
log 5
(
3
x y2
)
1
= 2 log x x + log x y − log x z
2
1
=
2 + log x y − log x z
2
(d)
log a xy 3
1
= log 5 3 x + log 5 y 2
=
(e)
= log a ( xy 3 ) 2
1
log 5 x + 2 log 5 y
3
1
log  
a
= log1 − log a
= − log a
1
= log a ( xy 3 )
2
1
1
log a ( x) + log a ( y 3 )
=
2
2
1
3
log a ( x) + log a ( y )
=
2
2
(f)
 8 
log 2  2 
 ab 
= log 2 8 − log 2 (ab 2 )
= log 2 23 − (log 2 a + log 2 b 2 )
= 3log 2 2 − log 2 a − 2 log 2 b
=
3 − log 2 a − 2 log 2 b
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5.
(a)
Write in terms of individual logs
3log 2 x − 1
(b)
3 − 2 log 3 x
= 3log 3 3 − 2 log 3 x
= log(2 x)3 − log10
3
(c)
8x
= log
10
4 x3
= log
5
log 5 x − 2
= log 3 33 − log 3 x 2
= log 3
27
x2
2 log 2 x + 3log 2 y + 5
(d)
= log 5 x − 2 log 5 5
= log 2 x 2 + log 2 y 3 + log 2 25
= log 2 ( 32 x 2 y 3 )
 x 
= log 5  
 25 
Topic: Solving Logarithmic and Exponential Equations
1.
(a)
Solve the following Logarithmic Equations.
log10 x = 2
(b)
log 5 x = −2
(d)
x = 5−2
1
x=
25
log 7 2 x = 2
2
x = 10
x = 100
(c)
log 4 x = 0.5
x = 40.5
x=2
2 x = 72
or 4
2 x = 49
x
=
(e)
log 2 ( x + 4 ) =
3
(f)
x+4=
23
x+4=
8
x=4
(g)
log 3 x 2 = 2
(h)
49
= 24.5
2
x
log 9   = −0.5
4
x
= 9−0.5
4
x
1
90.5
= =
4
9
x 1
=
4 3
4
x=
3
9
log 3 x = 2
x 2 = 32
x = 32
x2 = 9
x=9
x=
± 9=
±3
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log 2 ( x 2 + 2 x ) =
3
(i)
(j)
log x 4 = 4
4 = x4
x2 + 2x =
23
x= 44
x = 1.4142 to 4 d.p.
x2 + 2x − 8 =
0
( x + 4)( x − 2) =
0
x = −4 or 2
(k)
Checking; when x=-4, log28=3
when x=2, log28=3
log x 9 = 0.5
(m)
(l)
log x 64 = 2
9 = x 0.5
64 = x 2
x = 92
x = 81
x = ± 64
x =8
(n)
log x (2 x − 1) =
2
Bases cannot be negative
log x 3 = 2
3 = x2
2 x − 1 =x 2
0 = x2 − 2 x + 1
0= ( x − 1)
x − 1 =0
x =1
(o)
There are two answers of x=1
log x 10 = 3
10 = x
x= ± 3
x = 1.732
2
Bases cannot be negative
(p)
x = 10
x = 2.154
x=
3
(q)
2.
(a)
10log x = 4
10
=4
log 4 = log x Definition
x = 4 or
x=4
log x
(r)
change of base
eln( x + 4) = 7
x+4=
7
x=3
Solve for x in the Logarithmic Equations.
2 log x = log 4
(b)
log x = log 4
2
x2 = 4
x = 2 (-2 is not possible)
log( x + 5)= log x + log 6
log( x + 5) =
log 6 x
x+5 =
6x
5 = 5x
x =1
2 ln 2 x = ln 36
ln ( 2 x ) = ln 36
2
(c)
log 4 64 = x
log 64
log 4
x=3
3
4 x 2 = 36
(d)
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x2 = 9
x = 3 (-3 is not possible)
3log x = log 4
log x3 = log 4
x3 = 4
x
=
3
4 ≈ 1.587 to 3 d.p.
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(e)
2=
log x log 2 x + log 3
(f)
2 log 3 2 − log 3 ( x + 1) =
log 3 5
2
log 3 4 − log 3 ( x + 1) =
log 3 5
2
 4 
log 3 
 = log 3 5
 x +1 
4
=5
x +1
=
4 5x + 5
−1 =5 x
−1
x=
5
3log 3 2 + log 3 ( x − 2) =
log 3 5
log x = log 6 x
x = 6x
x2 − 6 x =
0
x( x − 6) =
0
x = 6 (0 is not possible)
(g)
log 4 x + log 4 6 =
log 4 12
(h)
log 4 6 x = log 4 12
log 3 23 + log 3 ( x − 2) =
log 3 5
6 x = 12
x=2
(i)
log8 ( x + 2) =2 − log8 2
log 3 8 ( x − 2 ) =
log 3 5
(j)
log 2 x + log 2 ( x + 4) =
5
log8 ( x + 2) + log8 2 =
2
(k)
8 x − 16 =
5
8 x = 21
21
5
x=
or 2
8
8
log 2 x =
5 - log 2 ( x + 4)
log8 2( x + 2) =
2
log 2 x( x + 4) =
5
2x + 4 =
64
2 x = 60
x = 30
25
x2 + 4x =
log x + log 3 =
log 5
log 3 x = log 5
3x = 5
5
x=
3
0
x 2 + 4 x - 32 =
( x - 4)( x + 8) =
0
(l)
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x=4
(-8 is not possible)
log x − log( x − 1) =
log 4
 x 
log 
 = log 4
 x −1 
x
=4
x −1
x 4x − 4
=
4 = 3x
4
x=
3
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(m)
log( x − 3) =
3
(n)
ln(4 - x) + ln 2 =
2 ln x
ln 2(4 - x) =
ln x 2
3
x −3 =
10
x −3 =
1000
8 - 2x =
x2
x = 1003
0 = x2 + 2x - 8
0=
( x + 4)( x - 2)
(o)
log 4 (2 x + 4) − 3 =
log 4 3
(p)
log 4 (2 x + 4) − log 4 3 =
3
 2x + 4 
log 4 
=3
 3 
2x + 4
= 64
3
2x + 4 =
192
2 x = 188
x = 94
3.
(a)
x = 2 (-4 is not possible)
2
log 2 x + 3log 2 2 =
log 2
x
2
log 2 x + log 2 23 =
log 2
x
2
log 2 8 x = log 2
x
2
8x =
x
2
8x = 2
1
4
1
1
x=
(- is not possible)
2
2
x2 =
Solve for y in terms of the other variables present.
=
y log x + log 4
log
log y = log 4 x
(b)
y = 4x
1
[log 5 + log x ]
2
1
log y = log 5 x
2
=
log y
1
=
=
(5 x) 2
log y log
5x
(c)
log
y log 4 − log x
=
4
log y = log
x
4
y=
x
(d)
y = 5x
log y + log x = log 4 + 2
log xy − log 4 =
2
 xy 
log   = 2
 4 
xy
= 102
4
xy = 400
y=
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5x
400
x
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(e)
log 5 x + log 5 y = log 5 3 + 1
(f)
=
2 ln
y ln e 2 − 3ln x
log 5 x + log 5 y − log 5 3 =
1
(g)
 xy 
log 5   = 1
 3 
xy
=5
3
xy = 15
15
y=
x
log 2 y + 2 log 2 x =
5 − log 2 4
ln=
y 2 ln e 2 − ln x 3
ln y 2 = ln
y
=2
25 x
y
= 72
25 x
y = 1225 x
2(log 4 y − 3log 4 x) =
3
log 7
x 2 y = 23
8
y= 2
x
(j)
ln y 3= 2 + ln x 3
2(log 4 y − log 4 x 3 ) =
3
ln y 3 − ln x 3 =
2
ln
y=
e2
x3
log 7 y − log 7 52 − log 7 x =
2
log 2 x 2 y = 3
3ln y= 2 + 3ln x
e2
x3
log 7 y = 2 log 7 5 + log 7 x + 2
(h)
log 2 y + log 2 x 2 =
5−2
(i)
y2 =
e2
x3
2
 y
log 4  3  = 3
x 
y2
= 64
x6
y 2 = 64 x 6
y3
=2
x3
y3
= e2
x3
y 3 = e2 x3
y = 8 x3
y = 3 e2 x3
(k)
2 y = ex
(l)
ln 2 y = x
y ln 2 = x
y=
x
ln 2
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10 y = 3x +1
log10 y = log 3x +1
y log10 = log 3x +1
y= ( x + 1) log 3
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1.
(a)
(b)
(c)
(d)
Solve the following exponential equations.
4 x = 20
4 x = 20
ln 4 x = ln 20 Take natural log of both sides
log 4 x = log 20 Take log of both sides
x ln 4 = ln 20 Using the third log law
x log 4 = log 20 Using the third log law
ln 20
log 20
x=
x=
ln 4
log 4
or
2.161
x
=
x = 2.161
1.6 x = 3.2
1.6 x = 3.2
log1.6 x = log 3.2
x log1.6 = log 3.2
log 3.2
x=
log1.6
x = 2.475
ln1.6 x = ln 3.2
x ln1.6 = ln 3.2
32 x = 0.125
32 x = 0.125
log 32 x = log 0.125
2 x log 3 = log 0.125
log 0.125
2x =
log 3
2 x = −1.8928
−1.8928
x=
2
x = −0.946
ln 32 x = ln 0.125
2 x ln 3 = ln 0.125
4.60.5 x = 100
4.60.5 x = 100
log 4.60.5 x = log100
0.5 x log 4.6 = log100
log100
0.5 x =
log 4.6
0.5 x = 3.018
3.0177
x=
0.5
x = 6.035
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ln 3.2
ln1.6
x = 2.475
x=
ln 0.125
ln 3
2 x = −1.8928
−1.8928
x=
2
x = −0.946
2x =
ln 4.60.5 x = ln100
0.5 x ln 4.6 = ln100
ln100
ln 4.6
0.5 x = 3.018
3.0177
x=
0.5
x = 6.035
0.5 x =
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(e)
6 x +1 = 40
6 x +1 = 40
log 6 x +1 = log 40
( x + 1) log 6 =
log 40
ln 6 x +1 = ln 40
( x + 1) ln 6 =
ln 40
ln 40
x +1 =
ln 6
x + 1 =2.0588
log 40
x +1 =
log 6
x + 1 =2.0588
=
x 2.0588 − 1
x = 1.0588
=
x 2.0588 − 1
x = 1.0588
(f)
(g)
(h)
11.22− x = 0.6
11.22− x = 0.6
log11.22− x = log 0.6
(2 − x) log11.2 =
log 0.6
log 0.6
2− x =
log11.2
2 − x =−0.2114
2 + 0.2114 =
x
x = 2.2114
ln11.22− x = ln 0.6
(2 − x) ln11.2 =
ln 0.6
ln 0.6
2− x =
ln11.2
2 − x =−0.2114
2 + 0.2114 =
x
x = 2.2114
2 x + 12 =
40 rearrange to get 2 x as the subject
2 x + 12 =
40 rearrange to get 2 x as the subject
x
2=
40 − 12
x
2=
40 − 12
2 x = 28
2 x = 28
log 2 x = log 28
x log 2 = log 28
log 28
x=
log 2
x = 4.8074
ln 2 x = ln 28
x ln 2 = ln 28
3 × 6 x − 10 =
12.5 rearrange to make 6 x the subject
ln 28
ln 2
x = 4.8074
x=
3 × 6 x − 10 =
12.5 rearrange to make 6 x the subject
3 × 6 x = 12.5 + 10
3 × 6 x = 12.5 + 10
3 × 6x =
22.5
22.5
6x =
3
x
6 = 7.5
3 × 6x =
22.5
22.5
6x =
3
x
6 = 7.5
log 6 x = log 7.5
x log 6 = log 7.5
log 7.5
x=
log 6
x = 1.1245
ln 6 x = ln 7.5
x ln 6 = ln 7.5
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ln 7.5
ln 6
x = 1.1245
x=
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Centre for Teaching and Learning
Numeracy
(i)
6 x + 4 = 611
as the bases are the same
the exponents can be equated
x+4=
11
x=7
(j)
5 x = 22 x +1
5 x = 22 x +1
log 5 x = log 22 x +1
5 (2 x + 1) log 2
x log=
log 2
=
x (2 x + 1)
log 5
=
x 0.4307(2 x + 1)
=
x 0.8614 x + 0.4307
1x − 0.8614 x =
0.4307
0.1386 x = 0.4307
0.4307
x=
0.1386
x = 3.108
2.
ln 5 x = ln 22 x +1
5 (2 x + 1) ln 2
x ln=
ln 2
=
x (2 x + 1)
ln 5
=
x 0.4307(2 x + 1)
=
x 0.8614 x + 0.4307
1x − 0.8614 x =
0.4307
0.1386 x = 0.4307
0.4307
x=
0.1386
x = 3.108
If an investor deposits $250 000 in an account that compounds at 7% p.a., how long is it
before this amount has increased to $400 000? (Compounded semi annually – half yearly).
Use the compound interest formula:=
A P(1 + i ) n
7% p.a. = 3.5% per half year
=
A P(1 + i ) n
=
400000 250000(1 + 0.035) n
400000
= 1.035n
250000
1.6 = 1.035n
log1.6 = log1.035n
log1.6
=n
log1.035
n = 13.66
It will take 13.66 half years for $250 000 to grow to $400 000. This is 6.83 years or, in reality,
7 years to the end of the period.
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