How do you calculate #log_4 28#?

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Answer 1 · 2016-08-24T22:00:23

#log_4 28=2.4037#

Using the identity #log_b a=(log_n a)/(log_n b)#, if we choose #n=10#, then

#log_4 28=log28/log4#

Now using tables as #log28=1.44716# and #log4=0.60206#

#log_4 28=1.44716/0.60206=2.4037#

Answer 2 · 2016-08-24T22:49:06

#2.404#

Log form and index form are interchangeable, and sometimes one form is easier to use than the other.

Note that: #log_a b = c" "harr" " a^c = b#

"The base stays the base and the other two change around"

#log_4 28 = x" "harr " " 4^x= 28#

We can see that this is not an integer answer because:

#4^2 = 16 and 4^3 = 64 # x is between 2 and 3.

#4^x= 28 " log each side"#

#xlog 4 = log 28" power law as well"#

#x = (log28)/(log4)" isolate x"#

#x = 2.404" calculator or tables"#

The same result would have been obtained using
the change of base law.