College Algebra (Abridged)

Let us solve an example using these rules of exponents. Simplify the following expression: (5 −1 6 −2 ) −3 (10 3 −3 ) 3 = 5 −1 ( −3 ) 6 ( −3 ) −2 ( −3 ) 10 1 ( 3 ) 3 ( 3 ) −3 ( 3 ) = 5 3 −18 6 10 3 9 −9 = 125 −18−9 6− ( −9 ) 1000 = −27 15 8 = 15 8 27 1.13 Logarithm Functions A logarithm function is the function that is written in the form of: ( ) = log , such that > 0 and ≠ 1 If = 2 , then ( ) = log 2 2 ( ) = Assuming different positive values of (it is necessary that 2 ( ) > 0) , log 2 ( ) ( , ) When we plot these values in the graph, we get the following graphical representation: 1 4 log 2 � 1 4 � = − 2 − 2 � 1 4 , − 2 � 1 2 log 2 � 1 2 � = − 1 − 1 � 1 2 , − 1 � 1 log 2 (1) = 0 0 (1,0) 2 log 2 (2) = 1 1 (2,1) 4 log 2 (4) = 2 2 (4,2) It should be noted that for all values of , (1) = 0 . Moreover, the value of ( ) will always be positive and greater than 0. The shape of the above graph becomes inverse (positive side only) whenever the values of a lies between 0 and 1.