Solve the following equation:

$$
\sqrt{0.12^{0.12 x}+5}+\sqrt{0.12^{0.12 x}+6}=5
$$
In the equation, by denoting the expression $0.12^{0.12 x}$, which appears twice, with $y$, we first solve the equation

$$
\sqrt{y+5}+\sqrt{y+6}=5
$$

By squaring, placing the resulting radical expression alone on one side, and squaring again, we get

$$
\begin{gathered}
2 y+11+2 \sqrt{(y+5)(y+6)}=25 \\
(y+5)(y+6)=y^{2}+11 y+30=(7-y)^{2}=49-14 y+y^{2} \\
y=0.76
\end{gathered}
$$

With this, the value of the left side is $2.4+2.6=5$, so the obtained value is a solution to equation (1).

Now, from the pure exponential equation $0.12^{0.12 x}=0.76$, calculating the approximate value of $x$ through the usual steps:

$$
\begin{gathered}
x=\frac{\lg 0.76}{0.12 \cdot \lg 0.12}=\frac{0.8808-1}{0.12(0.0792-1)}=\frac{0.1192}{0.12 \cdot 0.9208}, \\
\lg x-\lg 0.1192-(\lg 0.12+\lg 0.9208)=0.0328 \\
x=1.078, \quad \text { or } 1.079
\end{gathered}
$$

(After taking the logarithm twice, we only got accurate values for the first 3 digits.)

Lea Haáder (Székesfehérvár, Teleki B. lg. III. o. t.)