How do you solve #lnx=2#?

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Answer 1 · 2018-05-28T03:23:44

#x=e^2#

Base-#e# cancels out with the natural log (#ln#) function, so we can apply it to both sides. We get

#e^(lnx)=e^2#

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

Notice base-#e# and #ln# cancel, and we're left with

#x=e^2#

as our final answer.

Hope this helps!

Answer 2 · 2018-06-22T10:06:27

#color(green)(x = e^2 = 7.389#

#ln x = 2#

#color(green)(x = e^2 = 7.389#