Solve $(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$

Question

Find the real values of $x$ which satisfy the equation $(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$

My Attempt $$ e^{10}=e^{(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}}=e^{(5+2\sqrt{6})^{x^2-3}}\cdot e^{(5-2\sqrt{6})^{x^2-3}}=\bigg[e^{(5+2\sqrt{6})}\cdot e^{(5-2\sqrt{6})}\bigg]^{x^2-3}\\ =\Big[e^{10}\Big]^{x^2-3}=e^{10(x^2-3)}\\ \implies x^2-3=1\implies x=\pm2 $$ But my reference gives the solutions $x=\pm2,\pm\sqrt{2}$. Why am I missing the solutions $\pm\sqrt{2}$ in my attempt ?

Answer 1

Your third step is wrong: $$a^{b^x}\neq\left(a^b\right)^x.$$

Let $(5+2\sqrt6)^{x^2-3}=t$.

Thus, $t^2-10t+1=0,$ which gives $t\in\{5+\sqrt{24},5-2\sqrt6\}$.

Thus, $x^2-3=1$ or $x^2-3=-1.$

Answer 2

Hint: Note that $$(5+2\sqrt{6})(5-2\sqrt{6})=1$$