How do you calculate #log _10 (2)#?

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Answer 1 · 2018-03-10T17:43:42

#log_10(2)~~0.301#

Whenever we have a log problem like this, we can simplify using the Change of Base Property/Formula that states

#log_a(b)=(log_10(b)/log_10(a))#

Which is also equal to

#log_a(b)=ln(b)/ln(a)#, where #ln=#natural log

In our case, our base (#a#) is equal to #10#, and our #b# is #2#. Using the Change of Base Property, #log_10(2)# is equal to:

#=log_10(2)/(log_10(10))#

#=>ln2/ln10# (Which is the same thing as #log_10(2)#)

Which, through evaluating with a calculator, we get:

#log_10(2)~~0.301#