How do you solve #2lnx=1#?

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Answer 1 · 2018-03-18T09:29:32

#x = e^(1/2)#

Let's do PEMDAS backwards.

We don't have any addition or subtraction, so we can't really do anything there.

We have multiplication that we can undo to isolate the ln(x):
#2lnx = 1#
#lnx = 1/2 #

Now that the ln(x) is isolated, we can exponentiate:
#lnx = 1/2 implies e^(lnx) = e^(1/2) implies x = e^(1/2) #

our final answer.

Answer 2 · 2018-03-18T09:32:27

see below.

1) Divide each term by #2#
#ln(x)=1/2#

2) In solving #x#, we rewrite equation in logarithms form.

#e^(ln(x))=e^(1/2)#

3) Exponential and base-e log are inverse function.
#x=e^(1/2)#

4) The value of #x# is #e^(1/2)# or #1.6487(4d.p.)#