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Logarithms
What is a logarithm ?
Consider the expression 16 = 24
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An alternative, yet equivalent, way of
writing this expression is log2 16 = 4
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We see that the logarithm is the same as
the power or index in the original expression
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The two statements 16 = 24 log2 16 = 4 are
equivalent statements
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Example:
If we write down that 64 = 82 then the equivalent statement
using logarithms is log8 64 = 2
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(a)
(b)

(c)

(d)

(e)

(f)
Solution:
Now, the reality is that evaluating logarithms directly can be a very difficult process,
even for those who really understand them
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In that form we can usually get the answer
pretty quickly
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As suggested above, let’s convert this to exponential form
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However, most people can determine the exponent that we need on 4 to
get 16 once we do the exponentiation
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(b)
This one is similar to the previous part
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If you don’t know this answer right off the top of your head, start trying numbers
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In this case we
need an exponent of 4
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Changing the base will change the answer and so we always need
to keep track of the base
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(d)

Now, this one looks different from the previous parts, but it really isn’t any different
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First, notice that the only way that we can raise an integer to an integer power and get
a fraction as an answer is for the exponent to be negative
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Now, let’s ignore the fraction for a second and ask
case if we cube 5 we will get 125
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In this

So, it looks like we have the following,

(e)
Converting this logarithm to exponential form gives,

Now, just like the previous part, the only way that this is going to work out is if the
exponent is negative
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Example 2
Evaluate each of the following logarithms
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The final two evaluations are to
illustrate some of the properties of all logarithms that we’ll be looking at eventually
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because

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(d)

because

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(e)

because
that this one will work regardless of the base that we’re using
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Again, note that the

Example 3
Write each of the following as a single logarithm with a coefficient of 1
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Note as well that these examples are going to be using Properties 5 7 only we’ll be
using them in reverse
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(a) The first step here is to get rid of the coefficients on the logarithms
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In this direction, Property 7 says that we can move the
coefficient of a logarithm up to become a power on the term inside the logarithm
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We’ve now got a sum of two logarithms both with coefficients of 1 and both with the
same base
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Here is the answer for
this part
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We now have a difference of two logarithms and so we can use Property 6 in reverse
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Here is the answer to this part
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That isn’t a problem
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The reason for
this will be apparent in the next step
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The final topic that we need to discuss in this
section is the change of base formula
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However, that is about it, so what do we do if we need to evaluate another logarithm
that can’t be done easily as we did in the first set of examples that we looked at?
To do this we have the change of base formula
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where we can choose b to be anything we want it to be
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Here is the change of base formula
using both the common logarithm and the natural logarithm
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Solution
First, notice that we can’t use the same method to do this evaluation that we did in the
first set of examples
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If the 7
had been a 5, or a 25, or a 125, etc
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Therefore, we have
to use the change of base formula
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So, let’s use both and
verify that
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Now, let’s try the natural logarithm form of the change of base formula
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