Question

Solve for x. `log_x (log_3 x) = 2`?

Answer 1

The equation has no real solutions. However,
`x~~ 0.581 +- 0.669i` is a complex solution.

Explanation:

Using the formula

`log_b x = lnx/lnb`

we can rewrite our equation as

`ln(lnx/ln3)/lnx=2`

`ln(lnx/ln3) = ln(x^2)`

`ln(lnx/x^2 * 1/ln3) = 0`

We know that `ln 1 = 0`, hence

`lnx/x^2 *1/ln3 = 0 = > lnx/x^2=ln3`

`lnx = x^2ln3`

When we have an equation of this type, we want the arguement of the natural logarithm to have the same exponent as `x` in the right hand side (2, in this case). To do this, we must multiply both sides by two and use the property that:

`clog_b a = log_b a^c`

`:.ln(x^2)=2x^2ln3`

Raise `e` to both sides:

`x^2=e^(x^2 * 2ln3)`

Since `2ln3 = ln9`, we have

`x^2=9^(x^2)`

Let's solve a more general case. Consider the equation

`z=C^z`

for `z in CC`.

`z/C^z = 1`

`ze^(-zlnC)=1` `color(white)(ffffff) (E_1)`

The thought process here is that we wish to have everything in terms of `e` and the natural logarithm; this way, we can apply use the Lambert W function, which is defined as the inverse of the function

`f(w) = we^w`

`color(red)(W = f^(-1))`

As such, in the equation

`we^w = a`

we see that:

`a=f(w) => w=f^(-1)(a) = W(a)`

Now, multiply both sides of `E_1` by `-lnC`:

`-zlnCe^(-zlnC) = -lnC`

Let `z_0 = -zlnC`.

`z_0e^(z_0) = - lnC`

`color(blue)( => z_0 = W(-lnC))`

Finally, we have reached a solution! Undoing all substitutions, we get:

`-zlnC=W(-lnC) => color(blue)(z = (W(-lnC))/(-lnC))`

We have proven that the solution of the equation

`z=C^z` is

`z=(W(-lnC))/(-lnC)`

Our equation looked like this

`x^2=9^(x^2)`

`:. x^2=(W(-ln9))/(-ln9) => color(red)(x=sqrt((W(-ln9))/(-ln9)))`

This is the exact form of `x`. As it turns out, `x` is complex; its numerical value is approximately

`x ~~ 0.581 +- 0.669 i`