Prove that #2(log_10 5-1)=log_10 (1/4)#?
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#log (1/4)# does not equal #2(log_10
5-1)#
Lets start off by evaluating #2(log_10
5-1)# By simplifying the expression, we obtained #2log_10
4#
According to the properties of logarithmic function,
#log_bM^p = plog_b M#
Hence #2log_10
4# is equivalent to #log_10
4^2 or log_10 16#
Another property for logarithmic function is #log_bM = log_b N# if and only if M = N
As this equation shows,
#log (1/4)# #≠# #log 16 # as the values of #M# and #N# are not equal to each other.
Yes, #2(log_10 5-1)=log (1/4)#
Before we seek prove the identity, let us recall a few logarithmic relations.
#loga-logb=log(a/b)#, #mloga=log a^m# and #log_n n=1#
As from this we have #1=log_10 10#, we can write
#2(log_10 5-1)#
= #2(log_10 5-log_10 10)#
= #2(log_10 (5/10))#
= #2(log_10 (1/2))#
= #log_10 (1/2)^2#
= #log_10 (1/4)# or #log (1/4)#
As we do not write the base, when using #10# as base,
we have #log_10 (1/4)=log (1/4)#
