WHAT I HAVE DONE AND WHAT I WILL DO ABOUT ENGLISH FOR MATHEMATICS

Since I get English lesson in Junior High School, I have liked it. I interest in it very much. Nowadays I like English too, but it is not more than I like mathematics. First semester In university, I have got English lesson part I, with Mrs. Iin. We learn about English for mathematics. We have discussed about measurement and mathematics operation. Now, in second semester, I get English lesson part II with the lecture Mr. Marsigit. Firstly I think that English lesson is a difficult lesson. But after I get it I think studying is a funny activity. In English part II we learn about English for mathematics and mathematics education.
During I get English for mathematics lesson, I have done many activity like writing something In blog, doing the assignments and joining the English class. I always try my best to do the assignment which is the lecture asked. I always try my best to do the test. I also try so hard to comprehend the material which is given by the lecture although it is very difficult to comprehend it.
In the next time what I will do about English for mathematics? I think, I will try hard to develop the mathematics science. To make it come true, I must have curiosity. It will help me to make a scientific work or scientific paper. With scientific paper, I can participate the development of mathematics science. Not only it, I also must be a good mathematics teacher who can give the lesson to the student well. We know that a good mathematics teacher is a teacher who can control the classroom and create the atmosphere In order to make the student feel comfort and interest with mathematics lesson. A teacher is also as facilitator In guiding the student to find out the idea of the material. Because In the next time I will Be a mathematics teacher, I will try hard to be a good teacher in order to develop mathematics science. Keep fighting!!!

Exercise!!!!

1. The characteristics of logarithm
first characteristic,
Remember of Exponent function:
• a to the power of m times a to the power of n equals a to the power of m plus n
• a to the power of m over a to the power of n equals a to the power of m minus n
if b logarithm to the base a equals n, so b equals a to the power of n
if a logarithm to the base g equals x, so a equals g to the power of x
if b logarithm to the base g equals y, so b equals g to the power of y
what is the answer of a times b in bracket logarithm to the base g? if a logarithm to the base g equals x, so a equals g to the power of x and b logarithm to the base g equals y, so b equals g to the power of y, we can conclude a times b equals g to the power of x in bracket times g to the power of x in bracket, then we get a times b equals g to the power of x plus y. we can get a times b in bracket logarithm to the base g equals g to the power of x plus y in bracket logarithm to the base g, equals x plus y in bracket times g logarithm to the base g ( we know that g logarithm to the base g equals one), so it equals a plus b. we can conclude that a times b in bracket logarithm to the base g equals a plus b.
then we look:
• a over b equals g to the power of x in bracket over g to the power of y
• a over b equals g to the power of x minus y in bracket
• a over b logarithm to the base g equals g to the power of x minus y in bracket logarithm to the base g equals x minus y in bracket times g logarithm to the power of g, equals x minus y.
so we can conclude a over b logarithm to the base g equals a logarithm to the base g minus b logarithm to the power of g.
second characteristic,
• if a logarithm to the power of g equals x so a equals g to the power of x
• if b logarithm to the power of g equals y so b equals g to the power of y

Exercise!!!!

1. The characteristics of logarithm
first characteristic,
Remember of Exponent function:
• a to the power of m times a to the power of n equals a to the power of m plus n
• a to the power of m over a to the power of n equals a to the power of m minus n
if b logarithm to the base a equals n, so b equals a to the power of n
if a logarithm to the base g equals x, so a equals g to the power of x
if b logarithm to the base g equals y, so b equals g to the power of y
what is the answer of a times b in bracket logarithm to the base g? if a logarithm to the base g equals x, so a equals g to the power of x and b logarithm to the base g equals y, so b equals g to the power of y, we can conclude a times b equals g to the power of x in bracket times g to the power of x in bracket, then we get a times b equals g to the power of x plus y. we can get a times b in bracket logarithm to the base g equals g to the power of x plus y in bracket logarithm to the base g, equals x plus y in bracket times g logarithm to the base g ( we know that g logarithm to the base g equals one), so it equals a plus b. we can conclude that a times b in bracket logarithm to the base g equals a plus b.
then we look:
• a over b equals g to the power of x in bracket over g to the power of y
• a over b equals g to the power of x minus y in bracket
• a over b logarithm to the base g equals g to the power of x minus y in bracket logarithm to the base g equals x minus y in bracket times g logarithm to the power of g, equals x minus y.
so we can conclude a over b logarithm to the base g equals a logarithm to the base g minus b logarithm to the power of g.
second characteristic,
• if a logarithm to the power of g equals x so a equals g to the power of x
• if b logarithm to the power of g equals y so b equals g to the power of y