Answer to Question #168199 in Math for Festus

Question

If 10^x/2 + 10^-x/2 =4, prove that x = log(7 + 4√3)

Answer 1

Let $10^{x}=u, u>0.$ Then

$$u+u^{-1}=14$$

$$u^2-14u+1=0$$

$$u=7\pm4\sqrt{3}$$

$4\sqrt{3}=\sqrt{48}<7$

$u_1=7-4\sqrt{3},u_2=7+4\sqrt{3}$

$$x_1=\log{(7-4\sqrt{3})}$$

$$x_2=\log{(7+4\sqrt{3})}$$