Logarithm As Inverse Of Exponentiation
I. Understanding Operations And Their Inverses
So far we have seen how to raise a number to a given power, which is called exponentiation. We have also seen that for every mathematical operation there is an inverse operation, and that the inverse operation of exponentiation is $root$, that is, if:
$b^x = y$
$\Rightarrow b = \sqrt[x]{y}$
This method is useful when we are given $y$ and $x$ and need to find out $b$.
But what if we are given $b$ and $y$ and are required to find out $x$. In other words, how do we know the power to which $b$ must be raised to get $y$ as a result.
Hence, there is another type of inverse operation of exponentiation, which is called logarithm, and is denoted by the symbol $log$.
If,
$b^x = y$
$\Rightarrow \log_b{y} = x$
Now, given $b$ and $y$, we can answer the question $\unicode{0x201C}$To what power must $b$ be raised to get $y$ as a result$\unicode{0x201D}$.
Now, we will take a look at this inverse operation in the context of some other operations and their inverses.
Addition has two operands, and has only one inverse operation, that is, subtraction.
Similarly, multiplication also has two operands but has only one inverse operation, that is, division.
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But exponentiation, which also has two operands has two inverse operations, namely, root and logarithm.
Why is this difference? Let's try to answer this question.
We know that multiplication is commutative, that is, if $m \times n = p$, then $n \times m = p$ for all values of $m$ and $n$
Hence given $p$ and $m$ we can find out $n$ using the relationship: $n = p \div m$, or,
given $p$ and $n$ we can can find out $m$ using the relationship: $m = p \div n$
It is exactly the same reason why addition has only one inverse operation, because it is commutative.
However, exponentiation is not commutative, that is, for general values of $m$ and $n$:
$m^n \ne n^m$
That is why we need to inverse operations for exponentiation operation: $base^{exponent} = power$
One for the case where we know the $power$ and the $exponent$ and need to find the $base$ and the second one when we know the $power$ and the $base$ and need to find the $exponent$.
II. Definitions Related To Logarithm
Now we know that logarithm is the inverse operation of exponentiation, that is, given:
$b^e = p$
$\Rightarrow e = \log_b{(p)}$,
where $b$ is called the $base$ of the logarithm function.
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There are some common bases that we come across more often than others. These are as follows:
$\underline{Base\ 10}$ - when a base is not explicitly mentioned in a $log$ function it is taken as $10$, for example:
$\log{(x)}$ would mean logarithm of $x$ with $base = 10$, while
$\log_b{(x)}$ would mean logarithm of $x$ with $base = b$
$\underline{Base\ e}$ - $e$, also known as $Euler\ Number$ denotes an irrational number which is used as the base for natural logarithms, and has an approximate value of $2.71828...$
Natural logarithms are denoted by $ln$.
$ln(x)$ would mean logarithm of $x$ with $base = e$
$\underline{Base\ 2}$ - This is a common base for logarithms used in computer science, and is denoted by $lg$.
$lg(x)$ would mean logarithm of $x$ with $base = 2$
Note the following statement:
$\unicode{0x201C}Only\ positive\ numbers,\ other\ than\ 1\ can\ be\ the\ base\ of\ a\ logarithm.\unicode{0x201D}$
Let us try to understand why.
We will consider a base of $1$, and try to calculate $log_1{20}$
Let's say $log_1{20} = x$
$\therefore 1^x = 20$
Now, we know that no matter to what power you raise unity to, you will always get back $1$, and there is no such value of $x$ which will give you $20 = 1^x$. Therefore $1$ cannot be use as the base to calculate logarithm. This is the same reason why $0$ cannot be used as the base of a logarithm.
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To understand why bases cannot be negative, we can try to take a negative number, let's, say $-2$ and raise it to the power a rational number $\dfrac{m}{n}$..
$(-2)^{\frac{m}{n}} = a$
$\left(\sqrt[n]{-2}\right)^m = a$
Now, this can have a real solution only if $n$ is odd. It cannot have a real solution if $n$ is even. In that case the solution will be imaginary.
Hence, for real numbers, a negative base cannot be used for logarithms.
$\underline{IV.\ Logs\ That\ Do\ Not\ Exist}$
Let us try to answer the question:
$\unicode{0x201C}$For what value of $x$, do we get $b^x = 0$?$\unicode{0x201D}$
if $b \ne 0$ then there is no such value of $x$ which will satisfy the above condition, and if $b = 0$ there are infinitely many values of $x$ that will satisfy the above condition. Hence logarithm of zero is not defined.
$log_b(0) = undefined$
Now, let us try to answer another question:
$\unicode{0x201C}$For what value of $x$, do we get $b^x = a\ negative\ number$?$\unicode{0x201D}$
Let's take assume that $\log_b{n} = p$ where $b$ is a number greater than $0$ and not equal to $1$, and $n$ is a negative number.
Therefore, we have:
$b^p = n$
We know that $b$ is a positive number and raising it to any power, be it positive or negative will always result in a positive number. Hence $n$ can never be negative.
Hence, we can say that:
$\log_b{n} = undefined$ for $n \leqslant 0$
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IV. Special Cases Of Log:
For any valid base $b$, logarithm of the number to the same base is always $1$, that is:
$\log_{b}{b} = 1$ It is easy to see why, $b^1 = b$ $\therefore \log_{b}{b} = 1$
Logarithm of $1$ to any valid base is $0$.
This is also very simple to see, because any number $b$ to the power $0$ is always $1$.
$\therefore \log_{b}{1} = 0$
Now let us try the examples below:
$\underline{\text{Example 1}}$:
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$\underline{\text{Example 2}}$:
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$\underline{\text{Example 3}}$:
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$\underline{\text{Example: 4}}$:
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