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Logarithm

Welcome all visitors..
At this meeting, will be discussed about the material Logarithm. The basic understanding of logarithms is very simple, logarithm is an operation that looks for the value of a power.
It’s good you know in advance about the material properties of powers.

Next, we will recognize converting powers to logarithms, as follows:

if $\displaystyle a^b=c$ then $\displaystyle b=~^a\text{log }c$.
where $a$ is called the base.
Notes:
Especially for base 10 then it doesn’t need to be written, so it becomes “log” only.

Next we will be more familiar with seeing examples and even make as many examples of our own. Consider the following examples:

Convert the following exponents into logarithmic form.

1. $\displaystyle 2^3=8$
2. $\displaystyle 5^2=25$
3. $\displaystyle 3^4=81$
4. $\displaystyle 10^2=100$
5. $\displaystyle 10^5=100000$
6. $\displaystyle (fg-h)^{-cd^7y}=qr+2t^9$
Solution:
1. $\displaystyle 3=~^2\text{log }8$
2. $\displaystyle 2=~^5\text{log }25$
3. $\displaystyle 4=~^3\text{log }81$
4. $\displaystyle 2=\text{log }100$
5. $\displaystyle 5=\text{log }100000$
6. $\displaystyle -cd^7y=^{fg-h}\text{log}(qr+2t^9)$

That’s what logarithms are like, only changing the form of exponents. After this we will discuss the properties of logarithms. The nature of logarithms comes from the nature of powers and conversions to the logarithmic form itself. Immediately, here are the properties of logarithms:

Trait 1: $$^a\text{log }a^b=b$$

The proof for this property is simply to convert it to exponents. If we convert to exponents, we get:

$a^b=a^b$ ….. (Proven).

We move on to the next nature.
Trait 2: $$a^{^a\text{log }b}=b$$

We can prove this property by converting it to logarithmic form. So that we get:

$\displaystyle ^a\text{log }b=^a\text{log }b$ … (Proven).

Very simple isn’t it!. We move on to the next nature.

Trait 3: $$^a\text{log }b.~^b\text{log }c=~^a\text{log }c$$ We will prove this property as follows:
Suppose $\displaystyle ^a\text{log }b=m$ and $\displaystyle ^b\text{log }c=n$ then $\displaystyle a^m=b$ and $\displaystyle b^n=c$, substitute the b of the first equation into the second equation so that $\displaystyle \left(a^m\right)^n=a^{mn}=c$. So, $\displaystyle mn=~^a\text{log }c$ … (proven).

We move on to the next nature.
Trait 4: $$^a\text{log}(b.c)=~^a\text{log }b+~^a\text{log }c$$ Proof:
We change it to the form of exponents as follows: $$a^{^a\text{log }b+~^a\text{log }c}=b.c$$ $$a^{^a\text{log }b}.a^{^a\text{log }c}=b.c$$ from trait 2 then:
$b.c=b.c$ … (proven).

Trait 5: $$^a\text{log}\left(\frac{b}{c}\right)=~^a\text{log }b-~^a\text{log }c$$ Try to prove it yourself.

Exercise:
1. Prove that $\displaystyle \frac{^a\text{log }b}{^a\text{log }c}=~^c\text{log }b$
2. Prove that $\displaystyle ^a\text{log }b^n=n.~^a\text{log }b$
3. Prove that $\displaystyle ^a\text{log }b=\frac{1}{^b\text{log }a}$
4. Prove that $\displaystyle ^{a^n}\text{log }b=\frac{1}{n}.~^a\text{log }b$

Thus the material about this logarithm, thank you for visiting. See you in another post and hopefully useful.