Solving exponential equation of form $(a+b)^x + (a-b)^x = c$

Question

I'm given equation: $$\left(2-\sqrt3\right)^\frac{x}{2} + \left(2+\sqrt3\right)^\frac{x}{2} = 4,$$ and I'm stuck with it. Which way shall I dig into to solve it?

Answer 1

Let $u = (2+\sqrt{3})^{x/2}$. Using the observation that $2-\sqrt{3} = \frac{1}{2+\sqrt{3}}$, we have that $(2-\sqrt{3})^{x/2} + (2+\sqrt{3})^{x/2} = u+\frac{1}{u}$. We thus aim to solve $$ u + \frac{1}{u} = 4. $$ Since $u\ne 0$, this is equivalent to $u^2 - 4u + 1 = 0$. Using the quadratic formula, we obtain $u = 2\pm\sqrt{3}$. We thus have $$ (2+\sqrt{3})^{\frac{x}{2}} = 2\pm\sqrt{3} = (2+\sqrt{3})^{\pm 1}\implies \boxed{x = \pm 2}. $$

Answer 2

Using Kostiantyn's suggestion, we can rewrite your problem as: $$\frac{(2-\sqrt3)^\frac{x}{2}+(2-\sqrt3)^{-\frac{x}{2}}}2=2$$ We can then convert it into a hyperbolic trig function using the identity $a^x=e^{x\ln a}$. From there, use the inverse hyperbolic trig function to get your answer.