Formula iteration will solve this with a bit of cunning:

Rewrite the equation as:

n=ln[(2^55)/n]/log(2)

Then iterate:

n(r+1)=ln[(2^55)/n(r)]/log(2)

where n(r) is the r-th iterate.

Then with n(0)=1, we get:

n(1)=53,

n(2)=47.2721,

n(3)=47.4371,

n(4)=47.4321

n(5)=47.4322

n(6)=47.4322

With n=47.4322 the two sides of the original equation agree to 5 significant digits.

RonL