Solve the following equation:

$$
\left(\sqrt{a+\sqrt{a^{2}-1}}\right)^{x}+\left(\sqrt{a-\sqrt{a^{2}-1}}\right)^{x}=2 a
$$
To ensure that the expressions on the left side are meaningful in the realm of real numbers, we must ensure that the radicands of each square root are non-negative. From the inner square root, we have $a^{2}-1 \geqq 0$, which means either $a \leqq-1$ or $a \geqq 1$. In this case, $\sqrt{a^{2}-1}<\sqrt{a^{2}}=|a|$, so the number under the first (outer) square root is positive if $a \geqq 1$. Thus, the number under the second square root is also positive.

Notice that the bases of the two $x$-exponent powers are reciprocals of each other:

$$
\sqrt{a+\sqrt{a^{2}-1}} \cdot \sqrt{a-\sqrt{a^{2}-1}}=\sqrt{a^{2}-\left(a^{2}-1\right)}=1
$$

Based on this, if we denote the first term on the left side by $y$, the equation simplifies to:

$$
y+\frac{1}{y}=2 a
$$

from which we get

$$
y_{1,2}=a \pm \sqrt{a^{2}-1}
$$

the two roots are reciprocals of each other. Now,

$$
\begin{gathered}
\left(\sqrt{a+\sqrt{a^{2}-1}}\right)^{x}=y_{1}=a+\sqrt{a^{2}-1}-\text { so: } \\
x=\frac{\lg \left(a+\sqrt{a^{2}-1}\right)}{\lg \sqrt{a+\sqrt{a^{2}-1}}}=2
\end{gathered}
$$

assuming that the denominator is not 0, i.e., $a+\sqrt{a^{2}-1} \neq 1$; and similarly for $y_{2}=a-\sqrt{a^{2}-1}$, $x_{2}=-2$.

Indeed, with $x=2$, the value of the left side is

$$
\left(a+\sqrt{a^{2}-1}\right)+\left(a-\sqrt{a^{2}-1}\right)=2 a
$$

with $x=-2$, we need to take the reciprocals of these two terms, which, according to our observation, means that the two terms switch places, so $x=-2$ is also a root.

The excluded case follows if $a=1$; in this case, the equation is in the form $1^{x}+1^{x}=2$, which is satisfied by any $x$.

Ungár Tamás (Budapest, Toldy F. g. III. o. t.)