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Logarithms
The Law of Indices
Rules of indices are:
Indices
am × an = am+n
am ÷ an = am-n
(am )n = am×n
am/n = n am
a-m = 1/am
a0 = 1
Example
(a)
23  24 = 27
(add the indices)
(b)
105  102 = 103
(subtract the indices)
Definition of Logarithms
The logarithm to base a of a number, x, is the power to which you have to raise a to get x.
Example
The logarithm to base 2 of the number 8 is 3, since 23 = 8.
This is written
log2 8 = 3
Similarly
log216 = 4
log2 128 = 7
since 24 = 16
since 27 = 128
and
log10 1000 = 3
since 103 = 1000
Practice Questions
Evaluate:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
log10 100
log10 100000
log10 10
log525
log381
log464
log2 ½
log2 0.25
log10 0.1
log10 0.001
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Note that log10 10 = 1 since 101 = 10
In general, loga a = 1 since a1 = a
Another result to remember is loga 1 = 0
since a0 = 1.
E.g. log6 1 = 0 since 60 = 1
Practice Questions 2
Solve for x
E.g. log2 x = 5
This can be written as 25 = x, so x = 32
1. log10 x = 6
2. log7 x = 2
3. log2 x = 0
4. log10 x = -2
5. log10 x = 4
6. log2 x = -2
7. log4 x = -1
8. log10 x = 1
9. log10 x = 0
10. log3 x = -2
Theory of Logarithms
Example
Multiply 8  16
This is the same as 23  24 = 27 = 128
When we multiply powers of a number we add the indices.
Logarithms are a different way of writing powers of a number.
3
+ 4
= 7
log2 8 + log2 16 = log2 128
since log2 8 = 3 , log2 16 = 4 , log2 128 = 7
Adding logarithms of numbers is the same as multiplying the numbers.
Example2
100  10000 = 1000 000
102  104 = 106
2+4=6
log10 100 + log10 10 000 = log10 1000 000
In general,
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logax + loga y = loga xy
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Example log381 = 4, and 34 = 81
i.e. 3log3 81 = 81
alogaxy
= x  y
= alogax  alogay
so
loga xy = loga x + loga y
xy
HNC Engineering
or in general, alogax = x
Some results to remember
loga x = y means that ay = x
loga a = 1
loga 1 = 0
loga xy = loga x + loga y
loga(x/y) = loga x - loga y
loga xn = n logax
Practice questions
Write the answer in the form log x where x is a number
1. log 5 + log 2
2. 2 log 6
3. log 6 – log 3
4. ½ log 9
5. ½ log 16 + log 4
6. log 12 – 2 log 2 – log 9
7. 2 log 4 + log 9 – ½ log 144
Solve by taking logs to base 10
(e.g. 3x = 5
log 3x = log 5,
x log 3 = log 5,
1.
2.
3.
4.
5.
x= log5 / log 3
x = 1.465
2x = 1000 000
2x = 0.001
1.08x = 2
1.1x = 100
0.99x = 0.000 001
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Logarithms to base 10 and base e
We usually use base 10 or base e for our logarithms.
e is an irrational number which occurs naturally from the shape of the curve y = log x.
It has the value 2.718281828……
You can use base e in the same way that we have used the other bases above.
Practice questions
Apply the theory of logarithms to:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
Evaluate:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
y = px
log yz =
log x – log y =
ln yk =
a ln x + b ln y =
ey = x
ln xn – ln yn =
log (x/y) =
log (ab)c =
at = b
log2 4
log3 81
log10 0.01
log2 0.5
ln 1
ln e
ln 10
log381
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
log2 0.125
logx36 = 2 (Find x)
logx0.1 = -1 (Find x)
loxx 64 = 6 ( Find x)
log10x = 4 (find x)
log10 x = -3 (Find x)
log2 x = -4 (find x)
Solve the following equations using logs to base e
1.
2.
3.
4.
4y = 10
6y = 1
2x = 0.005
( 1 + r )n = 2 Find n when r = 10% or 0.1, then when r = 3% or 0.3
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ANSWERS
Practice Questions 1
1.
2
2.
5
3.
1
4.
2
5.
4
6.
3
7.
-1
8.
-2
9.
-1
10.
-3
Page 4
Theory
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
Practice Questions 2
1.
1 000 000
2.
49
3.
1
4.
0.01
5.
10 000
6.
0.25
7.
0.25
8.
10
9.
1
10.
1/9
Practice Questions 3
1.
log 10
2.
log 36
3.
log 2
4.
log 3
5.
log 16
6.
log 1/3
7.
log 12
1.
2.
3.
4.
5.
x = 19.93
x = -9.966
x = 9.006
x = 48.32
x = 1374.6
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logp y = x
log y + log z
log ( x/y)
k ln y
ln ax + ln yb = ln ( ax/yb )
ln x = y
ln ( xn / yn) = ln ((x/y)n) = n
ln (x/y)
log x – log y
c log ab = c log a + c log b
loga b = t
Evaluate
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)
(xi)
(xii)
(xiii)
(xiv)
(xv)
2
4
-2
-1
0
1
2.3026
4
-3
6
10
2
10 000
0.001
1/16 = 0.0625
Solve
1.
2.
3.
4.
1.661
0
-7.6439
7.2725;
23.45
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