Solve the equation

$$
e^{x}+2 e^{-x}=3
$$

where $e$ is the base of the natural logarithm (see the function table in your school textbook).
Our equation, when multiplied by $e^{x}$ and with the substitution $e^{x}=z$, transforms into

$$
z^{2}-3 z+2=0, \text { from which } z_{1}=1, z_{2}=2
$$

From $z_{1}=1$, it is obviously $x_{1}=0$. In the school function table (19th edition, 3rd o. "powers of e"), the value $e^{x}=2$ is bracketed by the data $e^{0,69}=1,9937$ and $e^{0,70}=2,0138$, which contain five significant digits. From this, instead of the usual interpolation to one additional digit, we can obtain two additional digits for $x$, the two-digit integer written with these is

$$
(20000-19937) \cdot \frac{100}{20138-19937} \approx 31
$$

thus, to four significant digits, $x_{2} \approx 0,6931$.

Pásztor István (Nyíregyháza, Vasvári P. g. II. o.t.)

Note. By calculating using base-10 logarithms, we can only determine three significant digits of $x_{2}$.