Processing Math: Done

From Förberedande kurs i matematik 1

Both left- and right-hand sides are positive for all values of x and this means that we can take the logarithm of both sides and get a more manageable equation,

LHSRHS=ln2−x2=−x2ln2=ln2e2x=ln2+lne2x=ln2+2xlne=ln2+2x1.

After a little rearranging, the equation becomes

x2+2ln2x+1=0.

We complete the square of the left-hand side,

x+1ln22−1ln22+1=0

and move the constant terms over to the right-hand side,

x+1ln22=1ln22−1.

It can be difficult to see whether the right-hand side is positive or not, but if we remember that e2 and that thus ln2lne=1, we must have that (1ln2)21, i.e. the right-hand side is positive.

The equation therefore has the solutions

x=−1ln21ln22−1

which can also be written as

x=ln2−11−(ln2)2.