How do you solve #5-lnx=7#?

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Answer 1 · 2016-10-03T03:47:40

#x=1/e^2#

Start off by subtracting #5# form both sides

#-lnx=2#

Now multiply both sides by negative #1#

#lnx=-2#

Rewrite both sides in terms of the base of the natural logarithm

#e^(lnx)=e^(-2)#

Using properties of logarithms and exponents we can write

#x=1/e^2#