In the second term on the right side, $17496=2^{3} \cdot 3^{7}$, therefore

$$\log _{\sqrt{18}} 17496 \sqrt{2}=\log _{\sqrt{18}} 2^{\frac{7}{2}} \cdot 3^{7}=\log _{\sqrt{18}}\left(2 \cdot 3^{2}\right)^{\frac{7}{2}}=\log _{\sqrt{18}} \sqrt{18^{7}}=7$$

In the other terms, with the notation $\lg x=y$, our equation simplifies to:

$$9^{2 y}+7=9^{y+1.5}-9^{y-1}=9^{y}\left(9^{1.5}-9^{-1}\right)=9^{y} \cdot 242 / 9$$

and further, with the notation $9^{2}=z$,

$$z^{2}+7=242 z / 9$$

From this, $z_{1} \approx 26.63$ and $z_{2} \approx 0.2629^{2}$. With these, from $\lg x=\lg z / \lg 9$,

$$\begin{aligned} & \lg x_{1} \approx 1.494 \\ & \lg x_{2} \approx(0.4198-1) / 0.9542=-0.6080=0.3920-1 \end{aligned}$$

and thus $x_{1} \approx 31.2, \quad x_{2} \approx 0.247$

Denes Marton (Budapest, Kölcsay F. Gymnasium III. grade)

Remark. Many contestants either did not perform the numerical calculation at all, or did so very superficially and carelessly. For example, one paper underlined a numerical expression as the final result, which still contained specified additions and square roots. There are also conceptual errors in the calculation of the logarithms of numbers between 0 and 1, and even primitive decimal point errors. - A calculation that is correct in principle but flawed in execution can cause a lot of trouble in practice.[^0]

[^0]: ${ }^{2}$ For the purpose of $z_{2}$, we calculated the square root of the discriminant to six significant figures without using logarithms. With logarithms, we can only get four significant figures, so the difference $242-237.3$, and thus $z_{2}$, would only have two significant figures. Our method allows us to get four significant figures for $z_{2}$ as well.