How do you calculate #log_2 4 - log_4 2#?

1 Answer
Aug 1, 2016

Answer:

#log_(2)4-log_(4)2# #=# #2-1/2# #=# #3/2#

Explanation:

When I write #log_(a)b# #=# #c#, in effect I ask to what power do I raise the base #a# to get #b#.

By this definition, if #log_(a)b# #=# #c#, then #a^(c)=b#.

Typically we use #"logs"# to the base #10# or the base #e#.

Can you tell me what is #log_(10)100+log_(10)1000# #?#