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

1 Answer
Oct 3, 2016

Answer:

#x=1/e^2#

Explanation:

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#