Logarithm Theory 2

 

1.      Given that Log 2 = 0.30101, evaluate the following,

i)        Log 16

Ii)  Log 128

 

                       Solution

i)         Log 16

 

= Log 2⁴

 

=   4 Log 2

Substitute 0^.30101 to Log 2

 

4        (0.30101) = 1.20404

 

ii)      Log 128        

= log 2⁷

=  7 Log 2

= 7 (0.30101)

= 2.10707

2.      Evaluate without any tables

3Log2 +Log20 – log1.6

               Solution

3Log2+Log20-Log 1.6

= Log2³+Log20-Log16/10

= Log{8x20x10/16}

= Log 100

= Log 10²

= 2Log10

= 2x1 =2