=±22
x
=
±
2
x
2
Using the fact that 2=ln2
2
=
e
ln
2
:
=±ln22
x
=
±
e
x
ln
2
2
−ln22=±1
x
e
−
x
ln
2
2
=
±
1
−ln22−ln22=∓ln22
−
x
ln
2
2
e
−
x
ln
2
2
=
∓
ln
2
2
Here we can apply a function known as the Lambert W function. If =
x
e
x
=
a
, then =()
x
=
W
(
a
)
.
−ln22=(∓ln22)
−
x
ln
2
2
=
W
(
∓
ln
2
2
)
=−2(∓ln22)ln2
x
=
−
2
W
(
∓
ln
2
2
)
ln
2
For negative values of
x
, ()
W
(
x
)
has 2 real solutions for −−1<<0
−
e
−
1
<
x
<
0
.
−ln22
−
ln
2
2
satisfies that condition, so we have 3 real solutions overall. One real solution for the positive input, and 2 real solutions for the negative input.
I used python to calculate the values. The dps property is the level of decimal precision, because the mpmath library does arbitrary precision math. For the 3rd output line, the -1 parameter gives us the second real solution for small negative inputs. If you are interested in complex solutions, you can change that second parameter to other integer values. 0 is the default number for that parameter.
