Download 1. Logarithms 1.1. Introduction. For this entire topic, when studying

1. Logarithms
1.1. Introduction. For this entire topic, when studying powers, we
assume that the base is positive i.e. when looking at ab , you can take it
that a is positive, a > 0. The powers can take any values, although we’ll
only be looking at negative and fraction values as we’ve studied those
in an earlier topic. So when discussing logarithms, 5−1/2 is something
we might see, but (−1)4 is not.
1.2. What are logarithms. The power rules for multiplication and
division allow us to do large calculations if the arguments are in the
same base e.g. 26 × 28 = 214 is easy, but 24 × 35 is difficult. Logarithms
are a method to represent a number in terms of it’s base and exponent
e.g. 64 can be written as 26 . This is described as follows, the logarithm
to base 2 of 64 is 6. It is sometimes described as, what power do I put
the base to, which gets me the given value.
logb (x) = p ↔ bp = x
So for the previous example log2 (64) = 6, because 26 = 64. Similarly
log10 (100000) = 5 as 105 = 100000.
We can use about powers to do calculations
1
−4
log3
3
= −4
=
log
3
34
7
log2 23
= log2 221 = 21
log5 53 × 57 = log5 510 = 10
Note that some logarithms do not take nice values. As an example,
log2 (10) is somewhere between 3 and 4, but it is not a fraction. It is
not even a surd, it is some other type of real number.
1.3. Log rules. Because of the power rules, we have associated logarithm rules.
logb (1) = 0
logb (x) + logb (y) = logb (xy)
logb (x) − logb (y) = logb (x/y)
p logb (x) = logb (xp )
We’ll prove one of these results here in the notes, one in a lecture, and
you’ll prove the other two. Let’s examine the rule logb (x) + logb (y) =
logb (xy). We’ll focus on the right hand side of the equation first. The
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expression logb (xy) is the power p such that bp = xy. Lets see if the
left hand side of the equation does the right thing.
blogb (x)+logb (y) = blogb (x) × blogb (y)
=x×y
So it satisfies the requirements to be logb (xy). This means logb (x) +
logb (y) is logb (xy)!
1.4. Translating logs. When taking logarithms, the different bases
will give different values e.g. looking at 912 , we could take the log to
base 9 and simply get
log9 (912 ) = 12
while taking the log to base 3 takes some more calculation
log3 (912 ) = log3 ((32 )12 ) = log3 (324 ) = 24
but is still a nice number, but taking the log to base 5 gives us
log5 (912 ) = 12 log5 (9)
and log5 (9) is not even rational.
Sometimes we can calculate the logarithm for one base, but not another. As an example, most calculators can only work with two bases,
either 10 or e.
The translation rule is
loga (b) logb (c) = loga (c)
You can see why this works by puting the common base a to the power
of the left hand side expression, and checking you get c. Hint: you’ll
need one of our power rules.
1.5. Relating variables using logs. Logarithms allow us to translate
products into sums and powers into products. This is useful when we
have know that there is a relationship between variables, but aren’t
sure precisely what that relationship is. Logarithms can help us to use
emperical data to determine how things are related.
1.5.1. Alice and Bob take samples. Suppose we know that x and y are
related by the equation y = 4 × 2x . However, Alice and Bob do not
know this equation, but they can take samples e.g. (1, 4), (2, 8) and
(5, 128).
By taking logarithms (ignoring bases for the moment), we can translate y = 4 × 2x into the equation log(y) = log(4) + x log(2). This looks
familar, setting Y = log(y) we have Y = log(2)x + log(4) which is a
linear equation. So by plotting log(y) against x, Alice and Bob see that
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the relationship is linear, and can reconstruct the orginal equation. Do
this, taking the log to base 2 which will make the values nicer.
1.5.2. Alice and Bob again. Suppose Alice and Bob are asked to find
the relationship between x and y in a different situation.
They take some samples and get the pairs (1, 10), (10, 10000) and
(100, 107 ).
By taking log10 of both variables, they translate the pairs into (0, 1),
(1, 4) and (2, 7). Remember this is for (log10 (x), log10 (y)).
They see that these sit on a straight line log10 (y) = 3 log10 (x) + 1
(we might write this as Y = 3X + 1 to hide the logs). To reconstruct
(what might be) the orginal equation, they put both sides as powers of
10, and find
10log(y) = 103 log(x)+1
y = 103 log(x) × 101
y = (10log(x) )3 × 101
y = 10x3
So Alice and Bob choose y = 10x3 as their equation.