Basic Logarithm tutorial

Question

anonymous · Answer

Starting of with some exponent laws:

$$a^1=a~~~8^1=8$$

$$a^0=1; a \neq 0 ~~~ \pi ^0 = 1$$

$$\frac{ a^m }{ a^n } = a ^{m-n}~~~\frac{ 81 }{ 9 }=\frac{ 3^4 }{ 3^2 }=3^{4-2}=3^2=9$$

$$a^m*a^n = a ^{m+n}~~~ x^4*x^2 = x ^{4+2}=x^6$$

$$(a^m)^{n} ~~~(7^2)^{3} = 7^{2*3}=7^6=117,649$$

$$a ^{\frac{ m }{ n }} = \sqrt[n]{a^m}~~~27^{\frac{ 2 }{ 3 }} = \sqrt[3]{27^2}=9$$

$$a ^{-n}=\frac{ 1 }{ a^n }~~~6^{-2}=\frac{ 1 }{ 6^2 }=\frac{ 1 }{ 36 }$$

$$(a*b)^n=a^n*b^n~~~(5*x)^3 = 5^3*x^3=125x^3$$

$$\left( \frac{ a }{ b } \right)^n=\frac{ a^n }{ b^n }~~\left( \frac{ 1 }{ 2 } \right)^3=\frac{ 1^3 }{ 2^3 }=\frac{ 1 }{ 8 }$$

$$\left( \frac{ a }{ b } \right)^{-n}=\left( \frac{ b }{ a }\right)^n=\frac{ b^n }{ a^n }~~~~~~~\left( \frac{ 2 }{ 3 } \right)^{-2}=\left( \frac{ 3 }{ 2 } \right)^2=\frac{ 3^2 }{ 2^2 }=\frac{ 9 }{ 4 }$$

rational · Answer

i think it is important to mention the domain of these laws when especially when introducing logarithms, for example :  $\large \left( \frac{ a }{ b } \right)^n=\frac{ a^n }{ b^n } $ doesn't work for all real numbers

anonymous · Answer

Using these exponent laws we can simplify questions as such:

$$\left( \frac{ 81x^2y ^{-3}z ^{\frac{ 1 }{ 2 }} }{ 243x^5y^6z ^{\frac{ 1 }{ 5 }} } \right)^{-2}$$

Looks complicated right? Well, lets use these laws and see how it works out to be.

$$81=3^4~~~243=3^5$$

$$= \left( 3^{-1}x ^{-3}y ^{-9}z ^{\frac{ 3 }{ 10 }} \right)^{-2}$$
$$= 3^2x^6y ^{18}z ^{\frac{ -3 }{ 5 }}\implies \frac{ 3^2x^6y ^{18} }{ \sqrt[5]{z^3} }$$

For such problems, you want to simplify your answers in two ways:
- With no negative or fractional exponents
- On a single line with fractional exponents

If you're interested try doing this one, and message me if you want the right answer :P

$$\frac{ \left( 4x^2y ^{-3}z \right)^{3/2} }{ \left( 8x^3y^4z ^{-3} \right) ^2}$$

Now for logarithms.

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$$y=2^x$$is an exponential function. The inverse of this function is $$x=2^y$$, and is a logarithmic function. You can see in the drawing, they are mirrors of each other, with the axis of symmetry being the diagonal line x = y.

Since they are inverses we can express the same information in different ways, but first we have to look at some conditions :P.

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|dw:1409550192394:dw|

I find the easiest way to remember how this relationship works is to learn to read the logarithm in english. 
You can see the base is the same for both.
$$\log_2 8 =3 $$ the exponent that I need to put on 2, in order to get an 8 is a 3.
$$\implies 2^3 = 8$$

This is a good technique to both convert between logarithms and exponents, and to evaluate simple exponents.

Exponent:
$$a^b = c$$
base a to the power of b equals c

and the inverse function;

Logarithms:
$$\log_a c = b$$
log base a of c equals b