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