# equation a^x=x

#### Vali

##### Member
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

##### Pessimist Singularitarian
Staff member
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:

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

Hence:

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

These imply:

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

And so:

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

#### Vali

##### Member
I would write:

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

Hence:

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

These imply:

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

And so:

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

##### Pessimist Singularitarian
Staff member
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:

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

And the first equation implies:

$$\displaystyle a^x=x$$

Hence:

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

#### Vali

##### Member
Thank you very much for your help!