How do you solve #-6+ln(3x)=0#?

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How do you solve #-6+ln(3x)=0#?

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Answer 1 · 2016-07-16T12:43:10

I found: #x=e^6/3#

Explanation:

We can start rearranging our expression:
#ln(3x)=6#
then use the definition of log to get (remembering that #ln# is the log in base #e#):
#3x=e^6#
so that:
#x=e^6/3#

or
#x=e^6/3=134.5#

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How do you solve #-6+ln(3x)=0#?

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