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How do you simplify #Log_(2)(10)#?

1 Answer

Aug 30, 2015

#log_2(10) = 1/log(2)#

Explanation:

Use the change of base formula: #log_a(b) = log_c(b)/log_a(b)#

#log_2(10) = log_10(10)/log_10(2) = 1/log(2)#

since #log_10 = log#

In general, #log_a(b) = log_b(b)/log_b(a) = 1/log_b(a)#

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