How do I solve for e ^ -ln2

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

Well, by definition: n^log(m, n) = m
where log(m, n) = logarithm of m to the base n.
So the answer you are looking at is 1/2 = 0.5
But if you want a proof of that then you could proceed like so.
(Note that ln2 = logarithm of 2 to the base e)

Let x = e^ -ln2
=> x = 1/(e^ln2)
=> e^ln2 = 1/x
=> ln(1/x) = ln2 ........ if x^y = z then log(z, x) = y i.e. log of z to base x = y by definition of logarithm
=> 1/x = 2
=> x = 1/2 = 0.5

Thus e ^ -ln2 = 1/2

The answer is 1/2 because e to the ln 2^-1 is e to the whatever e goes to to equal one half.

e^-ln2=1/e^ln2=1/2lne=1/2

e^ - ln2 = e^(2^-1) = 1/2