How do you calculate #Log_2 56 - log_4 49#?

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Answer 1 · 2016-08-09T05:20:53

#log_2 56-log_4 49 = 3#

Expression #=log_2 56-log_4 49#

First put the each term on the same base. I will choose #log_2#

We know that: #log_b x = (log_a x)/(log_a b)#

To change the base of #log_4 49# to #log_2#
We have #a=2, b=4 and x =49#

Hence: #log_4 49 = (log_2 49)/(log_2 4)#

#= (log_2 49)/2 = log_2 49^(1/2)#

#= log_2 sqrt(49) = log_2 7#

Therefore, Expression #= log_2 56 - log_2 7 = log_2 (56/7)#
#=log_2 8#

#= 3# (Since #8=2^3#)