How do you solve #ln3 + ln(4x) = 4#?

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Answer 1 · 2018-05-10T15:21:12

#=>x = e^4/12#

We need to make use of the logarithm properties:

#=> ln(a) + ln(b) = ln(ab)#

#=> e^ln(a) = a#

We start with

#ln(3) + ln(4x) = 4#

We combine the logarithms

#ln(3*4x) = 4#

#ln(12x) = 4#

We now raise both sides using base #e#

#e^ln(12x) = e^4#

#12x = e^4#

#=>x = e^4/12#