Solve the following equation by systematic trial:

$$
6^{x}+10^{x}=13^{x} .
$$
Our solution is based on the visualization of the exponential function graph in the textbook. The graph - if the base $a$ is greater than 1 - is a continuously rising line from left to right, with no "horizontal" segments, or in other words, for any $y > 0$, there is an $x$ such that $a^{x} = y$ (the logarithm of $y$ to the base $a$). We will also use without proof that the graph of the sum on the left side of the given equation has these same properties.

At $x=0,1,2$, the values on the left side are $2,16,136$, and on the right side, they are $1,13,169$. According to this, at the abscissa $x=1$, the graph on the left side is still higher, but at the abscissa $x=2$, the right side is already higher. Thus, the intersection point of the two graphs, whose abscissa is the root we are looking for, because its ordinate is the common value of the two sides, is between the lines $x=1$ and $x=2$.

At the intermediate point $x=3/2$, according to the table, rounded to two decimal places, on the one hand,

$$
6^{3 / 2}+10^{3 / 2}=\sqrt{216}+\sqrt{1000} \approx 14.70+31.62=46.32
$$

and on the other hand, $13^{3 / 2}=13 \sqrt{13} \approx 46.87$, which is larger than the left side, so the root is between 1 and $3/2=1.5$.

We will continue to calculate with logarithms. For simplicity, we will immediately calculate the difference between the right and left sides,

$$
f(x)=13^{x}-6^{x}-10^{x}
$$

and primarily consider its sign. If the difference is negative, we will try a larger $x$ in the next step, and if the difference is positive, we will try a smaller $x$. On the other hand, from the two $x$ values that give opposite signs of the difference, we will choose the next $x$ for the next attempt closer to the one where the absolute value of the difference is smaller. For $x=1.5$, the difference is around $+0.55$, and for $x=1$, it is -3. The latter has a much larger absolute value, so it is likely that $x$ is closer to 1.5 than to 1. Let's try $x=1.4$:

$$
f(1.4)=13^{1.4}-6^{1.4} \approx 10^{1.4} \approx 36.27-12.29-25.12=-1.14
$$

so $x$ is in the interval $(1.4 ; 1.5)$, closer to 1.5, because

$$
|+0.55|<|-1.14|
$$

Let's try $x=1.45$:

$$
f(1.45) \approx 40.55-13.20-27.35=0.00
$$

so $x$ is in the interval $(1.45 ; 1.5)$. Let's try $x=1.47$:

$$
f(1.47) \approx 43.62-13.98-29.65=-0.01
$$

and $x=1.473$:

$$
f(1.473) \approx 43.73-14.01-29.71=+0.01
$$

Considering that in reverse logarithmic calculations, we obtain the hundredths by interpolation, and even the multiples of the logarithms of the bases are rounded, we cannot refine the result further. Thus, the solution to the given equation, rounded to the thousandth, is $x=1.472$, or 1.473.

Gálfi István (Budapest, Kandó K. hír. ip. t. III. o. t.)

Note. The calculation is somewhat simpler if we write the equation in the form

$$
1.3^{x}-0.6^{x}-1=0
$$

by dividing the equation by $10^{x}$.

Horváth Sándor (Szeged, Radnóti M. g. II. o. t.)