How do you solve #log_3 32 = x#?

1 Answer
George C.
Nov 27, 2015

This is already 'solved' in that #x = log_3 32#

but you can use the change of base formula to find:

#x = log(32)/log(3) = ln(32)/ln(3) ~~ 3.15465#

Explanation:

The change of base formula tells use that if #a, b, c > 0# then:

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

This allows us to express #log_3 32# in terms of common (base #10#) logarithms or natural (base #e#) logarithms:

#log_3 32 = log(32)/log(3) = ln(32)/ln(3) ~~ 3.15465#

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