How do you solve #ln x = -3 #?

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Answer 1 · 2018-07-17T14:21:04

#x=1/e^3#

Given that

#\ln x=-3#

#e^{\ln x}=e^{-3}\quad (\text{raising powers to base e})#

#x=e^{-3}#

#x=1/e^3#

Answer 2 · 2018-07-19T23:34:57

#x=1/e^3#

We want to cancel out the natural log function. To go about doing this, we can apply its inverse, base #e# to both sides. We get

#e^lnx=e^-3#

This simplifies to

#x=1/e^3#

As a decimal, this is approximately #0.04979#

Hope this helps!