How do you solve #\ln ( x - 3) = 2#?

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Answer 1 · 2018-06-02T06:55:16

#x=10.39#

#In(x-3)=2#

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#log_e(x-3)=2#

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#x-3=e^2#

#x=3+e^2#

#x=10.39#

Answer 2 · 2018-06-03T02:18:03

#x~~10.39#

The key realization here is that #ln# and base #e# cancel each other out, because they are inverses of each other.

So to cancel out #ln# on the left, let's apply base #e# on both sides. We get

#e^ln(x-3)=e^2#

#cancel(e^ln)(x-3)=e^2#

#=>x-3=e^2#

Adding #3# to both sides, we now have

#x=e^2+3#

#x~~10.39#

Hope this helps!