The number of real solutions of the equation $$e^{4x} + 4e^{3x} - 58e^{2x} + 4e^x + 1 = 0$$ is ______

The equation is $$e^{4x} + 4e^{3x} - 58e^{2x} + 4e^x + 1 = 0$$.

Divide both sides by $$e^{2x}$$:

$$e^{2x} + 4e^x - 58 + 4e^{-x} + e^{-2x} = 0$$

$$(e^{2x} + e^{-2x}) + 4(e^x + e^{-x}) - 58 = 0$$

Let $$t = e^x + e^{-x}$$, where $$t \geq 2$$ (by AM-GM). Then $$t^2 = e^{2x} + 2 + e^{-2x}$$, so $$e^{2x} + e^{-2x} = t^2 - 2$$.

Substituting:

$$(t^2 - 2) + 4t - 58 = 0$$

$$t^2 + 4t - 60 = 0$$

$$(t + 10)(t - 6) = 0$$

Since $$t \geq 2$$, we have $$t = 6$$.

Now solving $$e^x + e^{-x} = 6$$:

$$e^{2x} - 6e^x + 1 = 0$$

$$e^x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = 3 \pm 2\sqrt{2}$$

Both values $$3 + 2\sqrt{2} > 0$$ and $$3 - 2\sqrt{2} > 0$$ are valid.

$$x = \ln(3 + 2\sqrt{2})$$ or $$x = \ln(3 - 2\sqrt{2})$$

Note that $$(3+2\sqrt{2})(3-2\sqrt{2}) = 1$$, so the two solutions are negatives of each other, confirming they are distinct.

Hence the number of real solutions is $$\boxed{2}$$.