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