EXERCISE IN ENGLISH PART 2 LESSON BY RIZAL AHMAD 08301244039 PMNR08

1. NATURES OF LOGARITHM
a to the power of m times a to the power of n equals a to the power of m plus n in bracket.
a to the power of m over a to the power of n equals a to the power of m minus n in bracket.
log base a of b equals n equivalent b equals a to the power of n
example :
question : log base g of a times b in bracket equals…
answer :
log base g of a equals x equivalent a equals g to the power of x
log base g of b equals y equivalent b equals g to the power of y
a times b equals g to the power of x in bracket times g to the power of y in bracket
a times b equals g to the power of x plus y in bracket
log base g of a times b in bracket equals log base g of g to the power of x plus y in bracket equals x plus y in bracket times log base g of g. Note : log base g of g equals 1. So, log base g of a times b in bracket equals x plus y. So, log base g of a times b in bracket equals log base g of a plus log base g of b.
a over b in bracket equals g to the power of x in bracket over g to the power of y in bracket, equivalent a over b in bracket equals g to the power of x minus y in bracket, equivalent log base g of a over b in bracket equals x minus y in bracket, equivalent log base g of a over b in bracket equals log base g of a minus log base g of b.
log base g of b to the power of n in bracket equals log base g of a times a times a times a times a no limit in bracket equivalent log base g of a to the power of n in bracket equals n times log base g of a.


2. HOW TO FIND THE VALUE FROM PHI NUMBER
Calculating approach phi can be elaborated as follows. Since is ahead known that by the ratio of area of circle to its distance square and ratio circle the circle with its diameter is constant. But, initially not yet been known that second of the constanta is of equal. Ancient book use the different constanta to be second of the ratio. Calculation Phi draw attention since pre-christian epoch ( about 1650 SM, in ancient Egypt). Since then hitherto a lot of all saintis conducting calculation of either through analytic and also by using computer. In modern epoch now calculation phi have time to be made by one of test to measure sophisticated and also a logarithm.
Searching number phi got from measurement of area of circle that is open bracket 8 over 9 times d close bracket to the power of 2; by d is diameter, and d equals 2 times r; by r is radius. So that got by equation of area of circle (A) equals open bracket 8 over 9 times 2 times r close bracket to the power of 2, equivalent A equals 64 over 81 times 4 times r to the power of 2, equivalent A equals 256 over 81 times r to the power of 2, equivalent 3,16 times r to the power of 2. Considering formula look for the area of circle equals phi times r to the power of 2, so, value of phi is 3,16.


3. HOW TO DETERMINE THE ABC FORMULA
To determine the ABC formula, we have had the common square equation, that is a times x to the power of 2 plus b times x plus c equals 0. Afterwards the equation is we alter in the form of perfect square equation, coefficient of x to the power of 2 turned into 1, so, the equation become x to the power of 2 plus open bracket b over a close bracket times x plus open bracket c over a close bracket equals 0. Than, left internode and joint right added with open square bracket b over open bracket 2 times a close bracket close square bracket to the power of 2, so, the equation become x to the power of 2 plus open bracket b over a close bracket times x plus open square bracket b over open bracket 2 times a close bracket close square bracket to the power of 2 plus open bracket c over a close bracket equals open square bracket b over open bracket 2 times a close bracket close square bracket to the power of 2. Than, alter x to the power of 2 plus open bracket b over a close bracket times x plus open square bracket b over open bracket 2 times a close bracket close square bracket to the power of 2 become open square bracket x plus open bracket b over 2 times a in bracket close bracket close square bracket to the power of 2, and move the internode c over a, so, the equation become open square bracket x plus open bracket b over 2 times a in bracket close bracket close square bracket to the power of 2 equals b to the power of 2 in bracket over 4 times open bracket a to the power of 2 close bracket plus c over a. than alter b to the power of 2 in bracket over 4 times open bracket a to the power of 2 close bracket plus c over a become open bracket b to the power of 2 minus 4 times a times c close bracket over open bracket 4 times open bracket a to the power of 2 close bracket. Than alter the mentioned equation become x plus open bracket b over 2 times a in bracket close bracket equals plus minus square root of open bracket b to the power of 2 minus 4 times a times c close bracket over 4 times a to the power of 2 close bracket. The last, move b over open bracket 2 times a close bracket to joint right, the equation become x equals negative b over 2 times a plus minus 1 over 2 times a times square root of open bracket b to the power of 2 minus 4 times a times c close bracket, so, x equals negative b plus minus square root of open bracket b to the power of 2 minus 4 times a times c close bracket over 2 times a.


4. HOW TO PROOF A SQUARE ROOT OF 2 IS IRRATIONAL NUMBER
To prove the square root of 2 is irrational number, provable through contradiction verification. Verification of through contradiction at ancient Greek era is verification that square root of 2 representing irrational number ( cannot be expressed by as integer comparison). This statement is provable by assuming on the contrary that square root of 2 is rational number, so that can be expressed by as comparison of integer a divided by b in bracket in simplest fraction. But if a divided by b in bracket equals square root of 2, hence a times 2 equals 2 times b times 2. This means a times 2 is even number. Because square from anomalous number not possible to be even, hence a is even number. Because a divided by b in bracket is simplest fraction of anomalous sure b ( even divided even fraction cause still can be made moderate). But because a is even number ( assume the its 2 times r meaning is a times 2 open bracket 4 times r times 2) is fold number 4, and b times 2 is fold number 2 ( even). Matter of this means b also represent the even number, and this represent the contradiction to previous conclusion that anomalous sure b. Because assumption of early that square root of two is rational result the happening of contradiction, the assumption wrong surely, and denied ( that square root of 2 is irrational) representing real correct statement.