How do you calculate log_4 28log428?

2 Answers
Aug 24, 2016

log_4 28=2.4037log428=2.4037

Explanation:

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

log_4 28=log28/log4log428=log28log4

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

log_4 28=1.44716/0.60206=2.4037log428=1.447160.60206=2.4037

Aug 24, 2016

2.4042.404

Explanation:

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 = blogab=c ac=b

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

log_4 28 = x" "harr " " 4^x= 28log428=x 4x=28

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

4^2 = 16 and 4^3 = 64 42=16and43=64 x is between 2 and 3.

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

xlog 4 = log 28" power law as well"xlog4=log28 power law as well

x = (log28)/(log4)" isolate x"x=log28log4 isolate x

x = 2.404" calculator or tables"x=2.404 calculator or tables

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