What are the real-number roots of Ln(x)=1/Ln(x)?

Question

I'm a bit doubtful. Am I suppose to multiply both side by Ln(x) and then square it? Then what am I suppose to the do with the "Ln" so that I get "x= the answer"? Please help me. Thanks.

Answer 1

If you multiply both sides by Ln(x), you get (Ln(x))^2 = 1, or Ln(x) = +1 or -1. Exponentiating, x = e or 1/e. It is necessary to verify that Ln(x) is not zero in any of this, and we see that this is so.

Answer 2

The Ln(x) means the natural logarithm of x. It is defined as:

If y = e^x then Ln(y) = x where e is the natural base e = 2.71828...

You can use this to solve your problem.

Ln(x) = 1/Ln(x)

(Ln(x))^2 = 1
Ln(x) = 1

Exponentiating: x = e^1 = e

Answer 3

ln x = 1/ln x
(ln x)² = 1
ln x = ±1
e^(ln x) = e^(±1)
x = e or 1/e

As TankAnswer noted, it can be either e or 1/e.

Answer 4

Oops.. hmm

Ln(X) ^ 2 = 1

ln(X) = 1 or -1

X = e or 1/e