A. Properties of Logarithm

1) a to the power of m times a to the power of n equals a to the power of m plus n in bracket.

2) a to the power of m over a to the power of n equals a to the power of m minus n in bracket.

3) Log b to the base of a equals n, means with b equals a to the power of n.

4) Log a to the base of g equal x, means with a equals g to the power of x.

5) Log b to the base of g equals y, means with b equals g to the power of y.

6) Log a times b in bracket to the base of g equals bla………bla……….bla……bla….

Example:

a. Log a to the base of g equals x, so a equals g to the power of x

b. Log b to the base of g equals y, so b equals g to the power of y.

a times b equals g to the power of x times g to the power of y.

a times b equals g to the power of x plus y in bracket.

Ø Log a times b in bracket to the base of g,

equals Log g to the power of x plus y in bracket to the base of g

equals x plus y in bracket times Log g to the base of g. Consider : Log g to the base g equals 1.

So, x plus y in bracket times Log g to the base of g equals 1.

So, Log a times b in bracket to the base of g equals Log a to the base of g plus Log b to the base of g.

Ø a over b equals g to the power of x over g to the power of y.

equivalen a over b equals g to the power of x minus in bracket.

equivalen Log a over b in bracket to the base of g equals Log g to the power of x minus y in bracket to the base of g.

equivalen Log a over b in bracket to the base of g equals x minus y in bracket times Log g to the base of g.

equivalen Log a over b in bracket to the base of g equals x minus y.

equivalen Log a over b in bracket to the base of g equals Log a to the base of g minus Log b to the base of g.

So, Log a over b in bracket to the base of g equals Log a to the base of g minus Log b to the base of g.

Ø Log a to the power of n to the base of g.

Log a to the power of n to the base of g equals Log a times a times a times a up to n factor,

Equivalen Log a to the power of n to the base of g equals Log a to the base of g plus Log a to the base of g plus Log a to the base of g plus bla…..bla…..bla….. plus Log a to the base of g plus Log a to the base of g.

Equivalen Log a to the power of n to the base of g equals n times Log a to the base of g

So, Log a to the power of n to the base of g equals n times Log a to the base of g.

B. Square Root of 2 is Irrational Number

Square root of 2 is the first irrational number is recognizedby Yunani people. They to say that long of diagonal from a right triangle with long of 2 side to form angled, one of them is irrational, from Phytagoras Theorem to get that long of triangle hypotenuse is a square root of 2.

So, How to prove that square root of 2 is irrational number?????????

Prove:

Square root of 2 is knowed by Yunani people that’s plane and that numeral is irrational. Supposing square root of 2 is rational, so must to be integer of x and y, so that square of 2 equals xper y, where y not zero. We can to confirm that x and y are not even (minimal one of them is a odd), because if both of them are even, we can to simplifying. In other meaning x per y we take the simplest. We get that x square per y square equals 2 or x square equals 2 times y square with other meaning x square is even. Because x square is even, so x is even, because that to be h even, so x equals 2 times h and thus

x square equals 2 times y square.

ð Open bracket 2 time h close bracket square equals 2 times y square.

ð 4 times h square equals 2 times y square,

ð y square equals 2 times h square.

Because y square is even, so that y is even too. This contradiction with we estimate that x and y cannot even, minimal one of them is odd, so that estimate is fllen and square root of 2 is irrational number.

(PROVED)

Prove with Math Logical, we use a implication.

If square root of 2 is irrational so that integer of x and y not zero, thus square root of 2 equal x per y. We are simboled square root of 2 is rational with p and statement of integer of x per y not zero so square root of 2 equals x per y with q, thus implication can be written with pèq, thus q wrong or –q right, because –qè-p equivalen with pèq, thus we get conclusion that p is wrong or square root of 2 is irrational number.

C. abc formula

a times x square plus b times x plus c equals zero is quadratic equation, so to find a roots of quadratic equation can search with abc formula.

This formula:

a times x square plus b times x plus c equals zero then both of articulations is overed with a.

Become:

x square plus b over a in bracket times x plus c over a equals zero.

Then c over a is transferred to right articulations and both of articulation plus with b over 2a in bracket square.

Become:

x square plus b over a in bracket times x plus b over 2a in bracket square equals b over 2a in bracket square minus c over a.

Then left articulation we can changed to perfect quadratic, become:

x plus b over 2a all in bracket square, then become:

x plus b over 2a all in bracket square equals b square over 4 times a square in bracket minus c over a, then right articulation equation of denominator become:

b square minus 4ac all over open bracket4 times a square close bracket, then become:

x plus b over 2a all in bracket square equals b square minus 4ac all over open bracket 4 times a square close bracket. Then both of articulation we root become:

x plus in bracket b over 2a close bracket equals plus minus square root of b square minus 4acall over open bracket 4 times a square close bracket.

Then we transfer b over 2a to right articulations, become:

x equals minus b over 2a plus minus square root of open bracket b square minus 4ac close bracket all over 2a.

So, x equals minus b plus minus square root of open bracket b square minus 4ac close bracket all over 2a.

D. Point of intersection

Find intersection of y equals x square minus 1 and y square plus x square equals 30.

Solving:

y square plus x square equals 30 is a circle has a radius of square root of 30 and centered in point (0,0) and y equals x square minus 1 is quadratic equation has vertex of -1. For to find a point of intersection we must to substitution y equal x square minus 1 to equation of y square plus x square equals 30.

è y equals x square minus 1 become y plus 1 equals x square then this equation we substitution to y square plus x square equals 30, become:

y square plus y plus 1 equals 30, then 30 we transfer to left articulation become:

y square plus y plus 1 minus 30 equals zero

y square plus y minus 29 equals zero

Then, we find y variable with abc formula, become:

y equals minus 1 plus minus square root of open bracket 1 minus 4 times 1times minus 29 close bracket all over 2

y equals minus 1 plus minus square root of 117 all over 2,

so, y₁ equals minus 1 plus square root of 117 all over 2

y₁ equals minus 1 plus 10 point 81 all over 2 equals 9 point 81 all over 2 equals 4 point 905

y₂ equals minus 1 minus square root of 117 all over 2

y₂ equals minus 1 minus 10 point 81 all over 2 equals minus 5 point 905

For y₁ equals 4 point 905, we substitution to x square equals y plus 1, become:

x square equals 4 point 905 plus 1

x square equals 5 point 905

x equals plus minus 2 point 43

For y₂ equals minus 5 point 905, we substitution to x square equals y plus 1, become:

x square equals minus 5 point 905 plus 1

x square equals minus 4 point 905, because x square minus so this not valid.

So, point of intersection are 2 point 905 comma 4 point 905 and minus 2 point 905 comma 4 point 905.