The equation
log(x+11)=((1/2))^{x+11}
has a unique solution, which cannot be found exactly (or guessed ... :) )
The existence is proved by Darboux-type reasoning:
for x=-10: log1=0<(1/2)^1=1/2
for x=-1: log10=1>(1/2)^10
Unicity is proven by monotonicity ... :)
The unique solution may be found numerically to be x=-9.5559
Darboux-type reasoning refers to:
f(x1)<0 and f(x2)>0 and f() continuous then there is at least one value x0 between x1 and x2 such that f(x0)=0