How do you simplify #log_10 5#?

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Answer 1 · 2015-09-14T06:29:47

It's not really possible to 'simplify' it, but it can be helpful to note:

#log_10 5 = 1 - log_10 2 ~~ 1 - 0.30103 = 0.69897#

#log_10(5) = log_10(10/2) = log_10(10) - log_10(2)#

#= 1 - log_10(2)#

So if you know #log_10(2) ~~ 0.30103#, then you don't need to memorise #log_10(5) ~~ 1 - 0.30103 = 0.69897#

You can also use the change of base formula:

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

to express #log_10(2)# in terms of natural logarithms:

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