For
, there exists values of
and
such that
. Find the value of
to the nearest integer.
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The equation can be written as x^(x+1) = (x+1)^x. A solution to this equation will also be a root of f(x) = ln(x) / x - ln(x+1)/(x+1). Since f(1) < 0 and f(e) >0, there is a root between 1 and e. Using Newton's Method with an initial guess of 2 converges quickly.