Prove that #2(log_10 5-1)=log_10 (1/4)#?

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Answer 1 · 2017-01-22T07:07:40

#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.

Answer 2 · 2017-01-22T07:37:50

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)#