Logarithms: definition and properties

It is known that $5^{3} = 125$ , but what happens in case that the unknown is the exponent? $5^{x} = 125$

In the previous example, it is enough to multiply $5$ by itself until we obtain $125$ . $5 \cdot 5 \cdot 5 = 125$

After multiplying $5$ three times, $125$ is obtained, so the value of the exponent is $3$ .

In the following example:

$3^{x} = 2187$

$3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 \cdot = 2.187$

So the exponent of 3 to obtain $2.187$ is $7$ .

There is a more practical way of finding out the exponents without having to multiply until finding the number: the logarithms.

In the first example $5^{3} = 125$ , if we apply a logarithm, we obtain the following expression: $l o g_{5} 125 = 3$ where $5$ is the base of the logarithm (as it was in the power), and the expression is read as logarithm of $125$ to base $5$ .

If we apply logarithms in the second example: $l o g_{3} 2.187 = 7$

Namely logarithm of $2.187$ to base $3$ .

Bearing in mind that the general expression of a power is $a^{n} = x$ the general expression of a logarithm is: $l o g_{a} x = n$

This expression allows us to calculate the number $n$ to which the number $a$ must be raised in order to produce the number $x$ .

It is only possible to calculate the logarithm of a positive number $> 0$ and its base must be $> 0$ and not equal to $1$ .

Example

$l o g_{3} 0$

It is not possible to express $0$ as a power of $3$ . In fact, there is no such number that multiplied by himself results in $0$ , therefore it is not possible to calculate.

Example

$l o g_{1} 20$

There is no way of expressing $20$ as a power with base $1$ because $1^{n} = 1$

Raising a number to $1$ does not really make sense, therefore it makes no sense to calculate the logarithm to base $1$ . We can deduce, therefore, that the base of a logarithm has to be a number greater than $1$ .

But, if it is only possible to calculate the logarithm of a number $> 0$ , does the logarithm of $1$ exist?

$l o g_{2} 1$

If we express $1$ as a power of base $2$ :

$l o g_{2} 1 = l o g_{2} 2^{0}$ since $2^{0} = 1$

For this reason $l o g_{2} 1 = l o g_{2} 2^{0} = 0$

The example allows to deduce that, in the general expression of a logarithm $l o g_{a} x = n$ , when $x = 1$ , the value of the logarithm, no matter its base, it will always be $0$ , since the only exponent to which it is possible to raise a number to obtain $1$ is $0$ . In other words, since: $a^{0} = 1$ then $l o g_{a} 1 = 0$ .

Calculating simple logarithms can be done immediately if we express the value of $x$ as a power of the same base as the logarithm.

Example

Continuing with the initial example: $l o g_{5} 125 = l o g_{5} 5^{3} = 3$ So, $3$ is the number to which it is necessary to raise $5$ to obtain $125$ .

More cases: $l o g_{2} 4 = l o g_{2} 2^{2} = 2$ So that $2$ is the number to which it is necessary to raise $2$ to obtain $4$ .

$l o g_{10} 1.000 = l o g_{10} 10^{3} = 3$

Therefore $3$ is the number to which it is necessary to raise $10$ to obtain $1.000$ .

These examples introduce one of the properties of the logarithms: $l o g_{a} x^{y} = y \cdot l o g_{a} x$

But the logarithm of $a$ to base $a$ is always $1$ .

Example

$l o g_{2} 2 = 1$ because the number to which it is necessary to raise $2$ to obtain $2$ can be only $1$ .

So that $l o g_{a} a^{n} = n \cdot 1 = n$

Before reaching the exercises, it is necessary to remember that, being related to the powers, the logarithms are also related with the roots, since:

$\sqrt[n]{a} = a^{\frac{1}{n}} = x$

Then, in this case:

$l o g_{a} x = \frac{1}{n}$