Recall one important viewpoint of logarithms: a logarithm is an exponent.
In particular, what is the number
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$\,\log_b x\,$?
Answer: It is the power that $\,b\,$ must be raised to,
in order to get $\,x\,$.
It follows that [beautiful math coming... please be patient] $\,\log_b x=y\,$ is equivalent to $\,b^y=x\,$.
The number $\,b\,$ in the expression
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$\,\log_b x\,$ is called the base of the logarithm.
We need to pause and determine the allowable bases for logarithms.
Then, we will study several important properties that logarithms exhibit.
Since $\,1\,$ raised to any power is $\,1\,$,
and since $\,0\,$ raised to any positive power is $\,0\,$,
we certainly don't want to allow the base to be $\,1\,$ or $\,0\,$.
For example, consider these dilemmas:
What is ‘$\,\log_1 5\,$’? There isn't any power that $\,1\,$ can be raised to, to give $\,5\,$.
Or, what is ‘$\,\log_1 1\,$’? The number $\,1\,$, raised to any power, is $\,1\,$. So, which number would we ‘choose’ as our answer?
Negative bases present similar difficulties.
We want
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$\,\log_b x = y\,$ to be equivalent to
$\,b^y = x\,$.
But, consider an expression like
$\,(2)^y \,$ (where the base is $\,2\,$).
Since you can't take
even roots of negative numbers,
we run into lots of problems.
For example,
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$\,(2)^1=2\,$, no problem.
However,
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$\,(2)^{1.01}=(2)^{101/100} = \root{100}\of{(2)^{101}}\,$ isn't even defined,
since it involves an even root of a negative number.
For these reasons, we only allow positive bases for logarithms, and we exclude the number $\,1\,$.
Now, a positive number, raised to any power, is always positive. (Think about this!)
For this reason, logarithms cannot act on negative numbers.
For example, what would ‘$\,\log_2 (3)\,$’ be?
There is no power that $\,2\,$ can be
raised to, in order to get a negative answer.
We are now ready to make things precise:
Logarithms have some beautiful simplifying properties, which make them extremely valuable.
They can take a multiplication problem, and turn it into an addition problem (which is much simpler)!
They can take a division problem, and turn it into a subtraction problem (which is much simpler)!
They can take an exponentiation problem, and turn it into a multiplication problem (which is much simpler)!
Precisely, we have:
Let [beautiful math coming... please be patient] $\,b\gt 0\,$, $\,b\ne 1\,$, $\,x\gt 0\,$, and $\,y\gt 0\,$.
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$\log_b\,xy = \log_b x + \log_b y$
The log of a product is the sum of the logs.
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$\displaystyle \log_b\frac{x}{y} = \log_b x  \log_b y$
The log of a quotient is the difference of the logs.
For this final property, $\,y\,$ can be any real number:
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$\log_b\,x^y = y\,\log_b x$
You can bring exponents down.
The last descriptive phrase, you can bring exponents down, is of course a bit loose.
It could be more correctly described as the log of a number raised to a power, is the
power, times the log of the number.
However, this is a bit long, and the shorter phrase seems to work well for students.
Here is a proof of the first property. The remaining two proofs are similar.
Let
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$\ \log_b x = u\ $ and
$\ \log_b y = v\,$.
Then, $\ b^u = x\ $ and $\ b^v = y\ $.
Renaming, we have:
[beautiful math coming... please be patient] $\log_b\,xy$  $=$  $\,\log_b\,b^u\,b^v$  (substitution) 
$=$  [beautiful math coming... please be patient] $\log_b\,b^{u+v}$  (law of exponents)  
$=$  $\,u+v\,$  (definition of logarithm)  
$=$  [beautiful math coming... please be patient] $\log_b x + \log_b y$  (substitution) 
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
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