- Thread starter
- #1

- Thread starter Vali
- Start date

- Thread starter
- #1

- Admin
- #2

I would write: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

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

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:

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

- Thread starter
- #3

Thank you for your response.I don't understand why x = 1/lnaI 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:

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

from a^xlna-1=0 => a^x=1/lna; why x = 1/lna ?

- Admin
- #4

The second equation implies: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 ?

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

And the first equation implies:

\(\displaystyle a^x=x\)

Hence:

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

- Thread starter
- #5