# 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:

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