Solving the Equation a^x=x: What is C?

[Vali]

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

 

[MarkFL]

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 would write:

$$f(x)=a^x-x=0$$

Hence:

$$f'(x)=a^x\ln(a)-1=0$$

These imply:

$$x=\frac{1}{\ln(a)}=\log_a(e)\implies a^x=e$$

And so:

$$\ln(a)=\frac{1}{e}\implies a=e^{\frac{1}{e}}$$

 

[Vali]

I would write:

$$f(x)=a^x-x=0$$

Hence:

$$f'(x)=a^x\ln(a)-1=0$$

These imply:

$$x=\frac{1}{\ln(a)}=\log_a(e)\implies a^x=e$$

And so:

$$\ln(a)=\frac{1}{e}\implies a=e^{\frac{1}{e}}$$

Thank you for your response.I don't understand why x = 1/lna
from a^xlna-1=0 => a^x=1/lna; why x = 1/lna ?

 

[MarkFL]

Thank you for your response.I don't understand why x = 1/lna
from a^xlna-1=0 => a^x=1/lna; why x = 1/lna ?

The second equation implies:

$$a^x=\frac{1}{\ln(a)}$$

And the first equation implies:

$$a^x=x$$

Hence:

$$x=\frac{1}{\ln(a)}$$

 

[Vali]

Thank you very much for your help!