Like their inverse operations, logarithms have certain properties which can be used to manipulate logarithmic expressions. Believe it or not, the identities of exponents and logarithms are related, and this relationship can help you to understand the rules. First and foremost, since exponents and logarithms are inverse operations:
For \(a, b > 0\):
$$a^{\log_{a}(b)} = b$$ $$\log_{a}(b^a) = b$$
These two identities are very important for understanding the relationship between exponential and logarithmic identities, and we will see why over time.
Another very basic identity is as follows:
Where \(a, b > 0\):
$$\log_{a}(\frac{1}{b}) = -\log_{a}(b)$$
This makes sense based on the graphs of logarithmic functions (try plugging in values for certain logarithmic functions), and it can be proven algebraically.
Both logarithms have a base \(a\), so undo the logarithm by taking the \(a\)th exponent of both sides:
$$a^{\log_{a}(\frac{1}{b})} = a^{-\log_{a}(b)}$$
The exponent and logarithm on the left side immediately cancel each other out:
$$\frac{1}{b} = a^{-\log_{a}(b)}$$
The right-hand side has a negative sign in front of the exponent, so convert it to an exponent that omits this:
$$\frac{1}{b} = \frac{1}{a^{\log_{a}(b)}}$$
The exponent and logarithm now cancel each other out:
$$\frac{1}{b} = \frac{1}{b}$$
Thus this identity is true. Of course we realize that \(b \neq 0\).
Consider these identities as you work through the problems that will proceed in later sections.
One of the most common ways to manipulate an expression with a logarithm is to convert a product inside a logarithm into a sum of logarithms or vice-versa. This is done with the following identity:
If \(a, b, c > 0\):
$$\log_{c}(a) + \log_{c}(b) = \log_{c}(ab)$$
A proof of this identity will be shown.
All of the logarithms are of base \(c\), so apply an exponent with base \(c\) to both sides of the equation (yes this is a valid operation; even if it looks strange, it is commonly done):
$$c^{\log_{c}(a) + \log_{c}(b)} = c^{\log_{c}(ab)}$$
The right-hand side is easily simplified because the exponent and logarithm have the same base:
$$c^{\log_{c}(a) + \log_{c}(b)} = ab$$
Now, the left-hand side is more difficult to simplify. We need to use an exponent law to break the exponent into a product of two:
$$c^{\log_{c}(a) + \log_{c}(b)} = c^{\log_{c}(a)} \cdot c^{\log_{c}(b)} = a \cdot b = ab$$
Therefore the identity holds.
A few examples of usage of this identity will be shown.
Example 1: Based on the identity just stated:
$$\log_{7}(4) + 1 = \log_{7}(4) + \log_{7}(7) = \log_{7}(4 \cdot 7) = \log_{7}(28)$$
Example 2: The identity can also be used in the reverse direction to rewrite \(\log_{12}(125)\), provided that we can factor \(125\). \(125\) factors as \(5 \cdot 5 \cdot 5)\):
$$\log_{12}(125) = \log_{12}(5 \cdot 5 \cdot 5) = \log_{12}(5) + \log_{12}(5 \cdot 5) = \log_{12}(5) + \log_{12}(5) + \log_{12}(5) = 3\log_{12}(5)$$
Example 2 also hints at another identity that will be introduced later. It is worth your time to guess what that identity may be.
Example 3: We can simplify \(3^{\log_{3}(2)} \cdot 9^{\log_{3}(4)}\) as follows: begin by converting the exponent with a base of \(9\) to one of base \(3\), to match the base of the logarithm within the superscript:
$$3^{\log_{3}(2)} \cdot (3^2)^{\log_{3}(4)} = 3^{\log_{3}(2)} \cdot (3^{\log_{3}(4)})^2 = 3^{\log_{3}(2)} \cdot 3^{\log_{3}(4)} \cdot 3^{\log_{3}(4)}$$
All of the exponents have the same base, so they can be combined via adding the superscripts together:
$$3^{\log_{3}(2) + \log_{3}(4) + \log_{3}(4)}$$
The logarithms now all have the same base, so we can condense them into one logarithm of base \(3\) by multiplying the inputs into the logarithms (also known as arguments):
$$3^{\log_{3}(2 \cdot 4 \cdot 4)} = 3^{\log_{3}(32)}$$
Now, the base of the exponent and logarithm are the same, so they cancel out:
$$3^{\log_{3}(32)} = 32$$
As a side note, this expression could also have been simplified by simplifying the first exponent first, then converting the base of the second to match the base of the logarithm. However, the solution above was chosen because it illustrates the logarithm sum identity.
Often a logarithm is in an inconvenient base, but it is possible to convert it to a more desirable base with an identity.
If \(a, b, n > 0\):
$$\log_{a^n}(b^n) = \log_{a}(b)$$
As with the other identities, this one will be proven, in the box below.
Take the \(a^n\)th exponent of both sides of the equation:
$$b^n = (a^n)^{\log_{a}(b)}$$
The right-hand side's exponent and logarithm are the same, so they cancel out, but the superscripts must be rearranged. To ensure that they can be rearranged, notice that the following two expressions are equivalent:
$$a^{\log_{a}{b}} \cdot a^{\log_{a}(b)} \cdot ... \cdot a^{\log_{a}(b)}$$ The expression is multiplied by itself \(n\) times: $$a^n \cdot a^n \cdot ... a^n$$ This time the expression is multiplied by itself \(\log_{a}(b)\) times
Therefore:
$$b^n = (a^{\log_{a}(b)})^n \Rightarrow$$ $$b^n = b^n$$
Here is a relatively simple example of this identity in use.
Example 4: The expression \(\log_{8}(343)\) can be rewritten as \(\log_{2}(7)\) because \(2^3 = 8\) and \(7^3 = 343\).
A logarithm can also have a quotient as an argument. Remember that every quotient is a product, because
$$\frac{a}{b} = a \cdot \frac{1}{b}$$
As a result, the identity for rewriting a logarithm with a quotient as an argument is similar to the identity used when the argument is a product. We use this identity:
If \(a, b, c > 0\), then
$$\log_{c}(\frac{a}{b}) = \log_{c}(a) - \log_{c}(b)$$
In other words, the quotient can be rewritten as a difference of two logarithms of the same base. We can use the identity for the sum of two logarithms to prove this identity, shown in the box below:
We consider the expression
$$\log_{c}(\frac{a}{b})$$
We can rewrite the argument as a product:
$$\log_{c}(a \cdot \frac{1}{b})$$
Now expand this product into sums:
$$\log_{c}(a) + \log_{c}(\frac{1}{b})$$
We know that \(\log_{c}(\frac{1}{b}) = -\log_{c}(b)\), so the expression becomes
$$\log_{c}(a) - \log_{c}(b)$$
and thus the identity holds.
The applications of this identity are similar to the previous one, and will be demonstrated with more examples.
Example 5: Consider the expression \(\log_{4}(8) - \log_{4}(2)\). We can divide the arguments since the bases are the same, giving
$$\log_{4}(4) = 1$$
Example 6: The difference-quotient identity can also be used to prove that \(\log_{a}(1) = 0\) where \(a > 0\). Prove this fact.
Solution: We consider a positive real number \(b\). Now we see that
$$\log_{a}(b) = \log_{a}(\frac{b}{1})$$
We can expand this logarithm into a difference of two logarithms:
$$\log_{a}(b) - \log_{a}(1)$$
However, we know that
$$\log_{a}(b) - \log_{a}(1) = \log_{a}(b)$$
So subtracting \(\log_{a}(b)\) from both sides of the equation gives
$$-\log_{a}(1) = 0 \Rightarrow$$ $$\log_{a}(1) = 0$$
Here is another example.
Example 7: Simplify the expression \(\log_{4}(12) - \log_{16}(24)\).
Solution:The base of the first logarithm can be converted to match the base of the second. Since \(16 = 4^2\), we can square the base if we square the argument. In general:
\(\log_{a^2}(b^2) = \log_{a}(b)\)
So we have
\(\log_{16}(144) - \log_{16}(24)\)
Now the bases are the same, we can divide the arguments, and get \(\log_{16}(6)\).
In general, a coefficient in front of a logarithm is preferred over an exponent inside a logarithm. There is another identity that can convert a logarithmic expression between these two forms:
If \(a, b, c > 0\):
$$\log_{c}(a^b) = b\log_{c}(a)$$
As usual, a proof of the identity will be given.
Take both sides of the equation to the \(c\)th exponent:
$$c^{\log_{c}(a^b)} = c^{b\log_{c}(a)}$$
The left side can be simplified immediately:
$$a^b = c^{b\log_{c}(a)}$$
The right-hand side's superscript is a product, and it needs to be broken up so only the logarithm remains:
$$a^b = (c^{\log_{c}(a)})^b \Rightarrow$$ $$a^b = a^b$$
Now let's move to some examples of where this identity can be used.
Example 8: Simplify \(\log_{4}(11^{43})\).
Solution: We can pull the superscript out of the exponent as follows:
$$\log_{4}(11^{43}) = 43\log_{4}(11)$$
This next example looks very similar to Example 8, but is more complicated.
Example 9: Simplify \(\log_{6}(36^{14})\).
Solution: Again, relocate the superscript outside of the exponent:
$$14\log_{6}(36)$$
However, \(36 = 6^2\), so
$$14\log_{6}(36) = 14\log_{6}(6^2) = 14 \cdot 2 = 28$$
Here is one more example that combines a few different identities. It can serve as a check-up to determine if you are understanding the identities presented in the article thus far.
Example 10: Simplify \(\log_{2}(2^3) + \log_{2}(96)\).
Solution: (note there are multiple forms of the final answer that can be considered "simplified")
Multiply the arguments, since the bases of the two logarithms are the same:
$$\log_{2}(4^3 \cdot 48)$$
The \(96\) has factors of \(2\), so factor out all the \(2\)s:
$$\log_{2}(4^3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 3) =$$ $$\log_{2}(4^3 \cdot 2^4 \cdot 3)$$
The \(3\) cannot be simplified further within a base \(2\) logarithm, so reverse the multiplication-addition identity to leave just powers of \(2\) and \(4\):
$$\log_{2}(4^3 \cdot 2^4) + \log_{2}(3)$$
Now, the exponents should have the same base. Convert \(2^4\) as follows:
$$2^4 = 2 \cdot 2 \cdot 2 \cdot 2 = 4 \cdot 4 = 4^2$$
Now both exponents have the same base:
$$\log_{2}(4^3 \cdot 4^2) + \log_{2}(3) = \log_{2}(4^5) + \log_{2}(3)$$
In the first logarithm, the superscript can be turned into a coefficient:
$$5\log_{2}(4) + \log_{2}(3)$$
Since \(4 = 2^2\), the expression becomes
$$5\log_{2}(2^2) + \log_{2}(3) = 5(2) + \log_{2}(3) = 10 + \log_{2}(3)$$
There is one final identity that needs to be covered because it is very useful, especially where you have products of logarithms. Following the usual restrictions on the variables:
$$\log_{a}(c)\log_{c}(b) = \log_{a}(b)$$
This can get you a logarithm in a different base or eliminate a logarithm with an unwanted base.
Example 11: We can simplify \(\log_{7}(4) \cdot \log_{4}(12)\) with the identity above, as \(\log_{7}(12)\).
Example 12: Simplify \(\log_{6}(12) \cdot \log_{144}(14)\).
Solution: The argument of the first logarithm can be made to match the base of the second by squaring the first' logarithm's base and argument:
$$\log_{36}(144) \cdot \log_{144}(14)$$
Now use the identity to write this as one logarithm:
$$\log_{36}(14)$$
Logarithmic identities are very powerful tools in the study of exponents and logarithms. These identities will be helpful in calculus and perhaps other math courses you will take later. It is also important to realize that in many problems, it takes more than one identity to simplify the expression given or otherwise solve the problem. That is why showing many different types of examples is important.
Reference, courtesy of
"The Art of Problem Solving Intermediate Algebra" Rusczyk, Richard and Crawford, Matthew