Question

f(x) = x + ln x, x > 0
Find the value of x such that f(x) = f^-1(x) using algebraic methods.

Answer 1

This is an interesting problem. What have you tried?

Answer 2

I tried this
f(x) = f^-1(x)
f^2(x) = x
x + ln x + ln (x + ln x) = x
ln x = ln (1 / (x + ln x))
x = 1 / (x + ln x)
x^2 + x ln x = 1
And I am stuck, don't even know if I am suppose to do ff(x).

Answer 3

This is excellent work.

x + ln x + ln (x + ln x) = x

You went somewhere else after this. It should be realized that you CANNOT solve explicitly for x with that bare 'x' and another in the logarithm.

The function: y = x + ln(x)
The inverse x = y + ln(y)

If we could solve for y in the inverse, that should lead to a solution. Since we can't do that, there is thinking to go...

It is important to realize that if \(f(x) = f^{-1}(x)\), we must be on the line x = y. Do you believe this?

Answer 4

Yes

Answer 5

Since x = y, we must have

y + ln(y) = x + ln(x)

This is from the definition of both the function and its inverse. Agreed?

Answer 6

Yes

Answer 7

Perfect. We know an expression for y. Let's use it.

(x + ln(x)) + ln(x + ln(x)) = x + ln(x)

Looks horrible, doesn't it?

Answer 8

I do not understand this part. Why is the back x + ln x instead of x?

Answer 9

y = x + ln(x) It was just a substitution.

Answer 10

Oh I see, thanks, I can solve it now.

Answer 11

Maybe. What's your plan?

Answer 12

Get it to ln x = 0
x = e^0 = 1

Answer 13

How did you get to that?

Answer 14

x = x + ln x

Answer 15

Fair enough. Just substituting x = y. Very good.

Just for practice, you can also get there this way.

ln(x + ln(x)) = 0

Then x + ln(x) = 1

The expression on the left is just y, so y = 1.

Unique answers don't care how you find them. I am always delighted to see it work two different ways!

Answer 16

Thanks.