The Exercise:
1. Characteristic of Logarithm
a. If a to the power of m times a to the power of n equal a to the power of m plus n in bracket.
b. If a to the power of m over a to the power of n equal a to the power of m minus n in bracket.
c. If logarithm of b to the base a equal n. Equivalent with b equal a to the power of n.
In other words, Logarithm of a to the base g equal x, equivalent with a equal g to the power of x and logarithm of b to the base g equal y equivalent with b equal g to the power of y.
For the exercise: Logarithm of a times b in bracket to the base g equal?
E.g., Logarithm of a to the base g equal x so a equal g to the power of x. Logarithm of b to the base g equal y so b equal g to the power of y. If a times b equal g to the power x times g to the power of y. If a times b equal g to the power of x plus y in bracket. For the note logarithm of g to the base g equal 1.
Logarithms of a times b in bracket to the base g equal logarithms of g to the power of x plus y in bracket to the base g. Equivalent with x plus y in bracket times logarithm of g to the base g. Equivalent with x plus y. To remember that logarithms of g to the base g equal one. So, the result is x plus y.
If logarithms of a times b in bracket to the base g equal logarithm of a to the base g plus logarithm of b to the base g. And if logarithm of a to the power of n to the base g equal logarithm of a times a times a….until n factor in bracket to the base g.
As well as logarithm of a to the power of n to the base g equal logarithm of a to the base g plus….plus a to the base g. From the above logarithm of a to the base g is n multiple with sum from respectively logarithm of a to the base g.
So, be obtained logarithm of a to the power of n to the base g equal n times logarithm of a to the base g.
Other evidence from nature if a over b equal g to the power of x in bracket over g to the power of y in bracket. Equivalent with a over b equal g to the power of x minus y in bracket. So that the principle: of considering the logarithm so logarithm of a over b in bracket to the base g equal logarithm g to the power of x minus y to the base g. Equivalent with logarithm of a over b in bracket to the base g equal x minus y in bracket times logarithm of g to the base g. Because logarithm of g to the base g equal one. So, logarithm a over b in bracket to the base g equal x minus y.
Meanwhile logarithm of a over b in bracket to the base g equal logarithm of a to the base g minus logarithm of b to the base g. That in accordance with the nature of two logarithm.


2. How do obtain the value of phi?
Phi is constant in mathematics which is the comparison that is always around to keep the diameter. And in ancient Egypt, the Moscow Papyrus and the Rhind Papyrus have found the value of phi of land in the circle is seen with the same square eight over nine times the middle line. Can be describe:
Note that area is eight over nine times diameter in bracket square.
With diameter equal two times r, r is radius.
So, the area equal eight over nine times diameter in bracket squared.
Six-ty four over Eighty one times four times r squared.
Equivalent with two hundred fifty-six over eighty one times r squared. So, the result is three point sixteen times r squared.
Meanwhile value of phi is a commonly used three point fourteen or twenty-two over seven but for the more precisely, have sought to more than
one billion two hundred and fourteen million one hundred thousand. decimal place. Phi values to ten decimal place is three comma one four one five nine two six five three five eight.
If proven in the following as an example:
If r=seven centimeter and the circumference of circle is forty-four centimeter.
So, phi value equal forty-four over two times seven in bracket. The result for phi is twenty-two over seven.


3. How do I get the formula a, b, c of quadratic equation?
If a times x squared plus b times x plus c equal nought.
Equivalent with a times x squared plus b times x equal minus c.
Equivalent with x squared plus b times x over all with a equal minus c over a.
Equal x squared plus b times x over a plus open bracket b over two times a close bracket square equal minus c over a plus open bracket b over two times a close bracket squared.
Equivalent open bracket x plus b over two times a close bracket squared equal b squared minus four times a times c in bracket over all four times a squared.
Equal with x plus b over two times a equal plus or minus squared root of b squared minus four times a times c over all squared root of four times a squared.
Equivalent x plus b over a equal plus or minus squared root of b squared minus four times a times c over all two times a.
So, x equal minus b plus or minus squared root of b squared minus four times a times c over all two times a.


4. Show that root of two is the irrational number?
E.g. the root of two is rational number in squared root of two equal a over b, where a and b as an integer prime number, so a equal squared root of two times b or a squared equal two times b squared.
Because a squared equal two times an integer number so a squared is even number so a is also even number. E.g. a equal two times c so the equation be:
four times c squared equal two times b squared.
two times c squared equal b squared.
So that b squared is even and b is also even. But even this is not possible because is it a relatively prime number. So the assumption that squared root of two is rational has brought us to the impossibility should be canceled.


5. How to search for intersection between y equal x squared minus one and x squared plus y squared equal thirty?
Note that y equal x squared minus one. Equivalent with x square equal y plus one.
To, come in equality x squares equal thirty. So, y plus one in bracket plus y squared equal thirty. y squared plus y minus twenty-nine equal nought.
Have known that a equal one, b equal one and c equal minus twenty-nine.
Thus, search for the value y by using the a, b, c formula:
y equal minus b plus or minus squared root of b squared minus four times a times c over all two times a.
So, minus one plus or minus squared root of one squared minus four times one times minus twenty-nine over all two times one equal minus one plus or minus squared root of one hundred seventeen over all two. Equal minus one plus or minus three times square root of thirteen over all two.
For y equal minus one plus three times squared root of thirteen over all two. So x equal squared root of one plus three times squared root of thirteen over all two.
And for y equal minus one minus three times squared root of thirteen over all two. So, the x equal squared root of one minus three times squared root of thirteen over all two.
Thus, the intersection point are squared root of one plus three times squared root of thirteen over all two point minus one plus three times squared root of thirteen over all two in bracket. And squared root of one minus three times squared root of thirteen over all two point minus one minus three times squared root of thirteen over all two in bracket.