If the equation a^x=x with a>1 has one solution then:
A)a=1/e
B)a=e
C)a=e^(1/e)
D)a=e^e
E)1/(e^e)
The right answer is C.I tried to derivate then to resolve f'(x) but didn't work
I did it with some help from another user
f(x)=a^x-x=0
f'(x)=a^xlna-1=0
a^xlna = 1 => a^x = 1/lna (from f'(x)=0)
But a^x=x (from initial equation) si x=1/lna=log_a(e)=>a^x=e => e=1/lna=> a=e^(1/e)