# 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 $125$. $5\cdot5\cdot5=125$

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

In the following example:

$3^x=2187$

$3\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3\cdot=2.187$

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: $$log_5 125=3$$ 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: $$log_3 2.187=7$$

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: $$log_a x=n$$

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 $1$.

$log_3 0$

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.

$log_1 20$

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?

$log_2 1$

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

$log_2 1=log_2 2^0$ since $2^0=1$

For this reason $log_2 1=log_2 2^0=0$

The example allows to deduce that, in the general expression of a logarithm $log_a x=n$, 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 $log_a 1=0$.

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

Continuing with the initial example: $$log_5 125=log_5 5^3=3$$ So, $3$ is the number to which it is necessary to raise $5$ to obtain $125$.

More cases: $$log_2 4=log_2 2^2=2$$ So that $2$ is the number to which it is necessary to raise $2$ to obtain $4$.

$log_{10} 1.000=log_{10} 10^3=3$

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: $$log_a x^y = y \cdot log_a x$$

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

$$log_2 2=1$$ because the number to which it is necessary to raise $2$ to obtain $2$ can be only $1$.

So that $$log_a a^n=n\cdot 1=n$$

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:

$log_a x=\dfrac{1}{n}$