Question

How do you solve `2lnx=1`?

Answer 1

`x = e^(1/2)`

Explanation:

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

see below.

Explanation:

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.)`