n0ka_in_EnglisH II

I. The Nature of Logarithm
a. a to the power of m times a to the power of n equals a to the power of m plus n in bracket.
b. a to the power of m over a to the power of n equals a to the power of m minus n in bracket
c. Logarithm base a b equals n, so b equals a to the power of n
From the third nature of logarithm, so we can make the other equality:
a. logarithm the base g a equals a, so a equals g to the power of x
b. logarithm the base g b equals y, so b equals g to the power of y
If we have logarithm the base g a times b in bracket equals x, so it means that logarithm the base g a times b in bracket equivalent a times b equals g to the power of x.
Example:
Showed: - logarithm the base g a equals x, so a equals g to the power of x (as the first equality)
- logarithm the base g b equals y, so b equals g to the power of y (as the second equality)
• we can find the problem solving of a times b with that equality
From the first and the second equality, we get that:
a equals g to the power of x
b equals g to the power of y
So, a times b equals g to the power of x times g to the power of y
Equivalent with a times b equals g to the power of x plus y in bracket (the first nature of logarithm that a to the power of m times a to the power of n equals a to the power of m plus n in bracket)
Equivalent with logarithm the base g a times b in bracket equals logarithm g g to the power of x plus y in bracket
Equivalent with x plus y in bracket times logarithm the base g g
Remembered that logarithm the base g g equals one
Equivalent x plus y in bracket times one
So, logarithm the base g a times b in bracket equals x plus y
And remembered the equality that x equals logarithm the base a and y equals logarithm the base g b
So, logarithm the base g a times b in bracket equals logarithm the base g a plus logarithm the base g b
• we can find the problem solving of a over b with that equality
from the first and the second equality, we get that:
a equals g to the power of x
b equals g to the power of y
So, a plus b equals g to the power of x plus g to the power of y
Equivalent with a over b equals g to the power of a minus b in bracket (the second nature of logarithm that a to the power of m over a to the power of n equals a to the power of m minus n in bracket)
Equivalent with logarithm the base g a over b equals logarithm the base g g to the power of x minus yin bracket
Equivalent with logarithm the base g a over b equals x minus y in bracket times logarithm the base g g
Remembered again that logarithm the base g g equals one
Equivalent with logarithm the base a over b equals x minus y
And remembered the equality that x equals logarithm the base a and y equals logarithm the base g b
So, logarithm the base g a over b equals logarithm the base g a minus logarithm the base g b

Logarithm the base g a to the power of n equals logarithm the base g times open bracket a times a times a times…times a to n factor close bracket.
Equivalent with logarithm the base g a to the power of n equals logarithm the base g a plus logarithm the base g a sum of logarithm the base g a until n factor
So, logarithm the base g a to the power f n equals n times logarithm the base g a

II. Find the Value of Phi
The area of a circle showed equals eight over nine times the diameter of circle in bracket square, and the volume of aright cylinder is equals the area of the base times the altitude of cylinder. So, we can explain it:
The circle area equals open bracket eight over nine in bracket times diameter close bracket square
We know that diameter of circle is equals two time the radius of circle. So, we can get:
The circle area equals open bracket eight over nine in bracket times two times the radius of circle close bracket square
Equivalent with sixty four over eighty one in bracket times four times r square (r as radius)
Equivalent with two hundred fifty six over eighty one in bracket times r square
Equivalent with tree point sixteen times r square
So, Egypt had found the value of Phi, it is tree point sixteen.

III. Find the Formula of abc to Solving the Problem of Mathematics
The general form of square equality is a times x square plus b times x plus c equals zero
Then divided all with a
So, a over a in bracket times x square plus b over a in bracket times x plus c over a in bracket equals zero
Equivalent with x square plus b over a in bracket times x equals negative c over a in bracket
We complete the equality with perfect square
So, equivalent with x square plus b over a in bracket times x plus open bracket b over two times a in bracket close bracket square equals negative c over a in bracket plus open bracket b over two times a in bracket close bracket square
Equivalent with open bracket x plus b over two times a in bracket close bracket square equals b square minus four times a times a in bracket all over four times a square
Equivalent with x plus b over two times a in bracket equals plus minus the square root of b square minus four times a times c in bracket all over four times a square
Equivalent with x equals negative b plus minus the square root of b square minus four times a times c in bracket all over two times a in bracket
The value of x is negative b plus minus the square root of b square minus four times a times c in bracket all over two times a in bracket
We ever call this pattern as abc pattern.

IV. The Square Root of Two is Irrational
If we will approve that the square root of two is irrational, firstly we instance the square root of two as rational. It means that the square root of two equals a over b which a and b as prime. So, the square root of two equals a over b
Equivalent with a equals b times the square root of two or a equals b square times two
Because a square equals a integer so, a is also even number
If we instance a equals two times c, o the equality to become:
Four times c square equals two times b square
Equivalent with two times c square equals b square
So, b square is even number and b is even number, too. But it is impossible, because a and b is imposable as even number because a and b is relative prime. So, the assumption that the square root of two is rational had bought us to the impossible anything and it must canceled. So, we can see that the square root of two is irrational has approved.

V. The Point Intersect from Two Equality
y equals x square minus one as the (first equality)
x square plus y square equals thirty (as the second equality)
The first equality:
y equals x square minus one
Equivalent with y square equals x square minus one in bracket square
Equivalent with y square equals x to the power of four minus two times x square plus one
The second equality:
x square plus y square equals thirty
Equivalent with y square equals negative x square plus thirty
The first and the second equality joined:
y square equals y square
equivalent with x to the power of four minus two times x square plus one equals negative x square plus thirty
equivalent with x to the power of four minus x square equals twenty nine
equivalent with x square times x square minus one in bracket equals twenty nine
so, x square equals twenty nine or x square minus one equals thirty
for x square equals twenty nine, so x equals plus minus the square root of twenty nine, it means that x equals the square root of twenty nine or x equals negative the square root of twenty nine.
For x square equals thirty, so x equals plus minus the square root of that, it means that x equals the square root of thirty or x equals negative the square root of thirty
We had got the value of the x, then for get the value of y we input the values of x to the equality. In here, we use the first equality to find the value of y.
For x equals the square root of twenty nine
The first equality:
y equals x square minus one
input x equals the square of twenty nine to the equality
so, y equals the square root of twenty nine in bracket square minus one
equivalent with y equals twenty nine minus one
equivalent with y equals twenty eight
So, the first intersect from two equality above is x equals the square root of twenty nine and y equals twenty eight

For x equals negative the square root of twenty nine
The first equality:
y equals x square minus one
input x equals negative the square of twenty nine to the equality
So, y equals negative the square root of twenty nine in bracket square minus one
equivalent with y equals twenty nine minus one
equivalent with y equals twenty eight
so, the second intersect from two equality above is x equals negative the square root of twenty nine and y equals twenty eight

For x equals the square root of thirty
The first equality:
y equals x square minus one
input x equals the square of thirty to the equality
so, y equals the square root of thirty in bracket square minus one
equivalent with y equals thirty minus one
equivalent with y equals twenty nine
so, the third intersect from two equality above is x equals the square root of thirty and y equals twenty nine

For x equals negative the square root of thirty
The first equality:
y equals x square minus one
input x equals negative the square of thirty to the equality
so, y equals negative the square root of thirty in bracket square minus one
equivalent with y equals thirty minus one
equivalent with y equals twenty nine
so, the fourth intersect from two equality above is x equals negative the square root of thirty and y equals twenty nine