How do you solve #log x=3#?

1 Answer
Sep 17, 2015

#x = 1000#

Explanation:

Remember the definition of logarithm, that is
If #log_b(a) = c# is true, then #a = b^c#, and if no base is explicitly put we always assume it's the base 10 (unless it's written as #lnx# in which case the base is the irrational number #e#)

#logx = 3 rarr x = 10^3 rarr x = 1000#