Solve the following equation:

$$
\log _{x} 10+2 \log _{10 x} 10+3 \log _{100 x} 10=0
$$
volna.

I. solution: The values $x=0, x=1, x=\frac{1}{10}, x=\frac{1}{100}$ are excluded from the outset, because in this case the left side is meaningless.

Using the identity

$$
\log _{a} b=\frac{\lg b}{\lg a}
$$

we get:

$$
\log _{x} 10=\frac{1}{\lg x}, \quad \log _{10 x} 10=\frac{1}{\lg 10 x}=\frac{1}{1+\lg x}, \quad \log _{100 x} 10=\frac{1}{\lg 100 x}=\frac{1}{2+\lg x}
$$

Substituting these values into our equation:

$$
\frac{1}{\lg x}+\frac{2}{1+\lg x}+\frac{3}{2+\lg x}=0
$$

which simplifies to

$$
3(\lg x)^{2}+5 \lg x+1=0
$$

from which

$$
\lg x_{1,2}=\frac{-5 \pm \sqrt{13}}{6}
$$

thus

$$
x_{1}=10^{\frac{-5+\sqrt{13}}{6}} \approx 0.5857 \quad x_{2}=10^{\frac{-5-\sqrt{13}}{6}} \approx 0.0368
$$

Since we performed identical transformations and none of the roots match the excluded values, the roots of our equation are necessarily these and only these.

Lackner Györgyi (Bp. V.. Bolyai János Textile Institute IV. o. t.)

# II. solution: Let 

$$
\log _{x} 10=y, \quad \log _{10 x} 10=z, \quad \log _{100 x} 10=u
$$

In this sense,

$$
\begin{gathered}
x^{y}=10 \\
x^{z}=10^{1-z} \\
x^{u}=10^{1-2 u}
\end{gathered}
$$

From (1),

$$
x=10^{\frac{1}{y}}
$$

From (2),

$$
x=10^{\frac{1-z}{z}}
$$

From (3),

$$
x=10^{\frac{1-2 u}{u}}
$$

Thus,

$$
10^{\frac{1}{y}}=10^{\frac{1-z}{z}}=10^{\frac{1-2 u}{u}}
$$

from which

$$
\begin{array}{rll}
\frac{1}{y}=\frac{1-z}{z}, & \text { hence } & z=\frac{y}{1+y} \\
\frac{1}{y}=\frac{1-2 u}{u}, & \text { hence } & u=\frac{y}{1+2 y}
\end{array}
$$

Substituting these values into the equation:

$$
y+\frac{2 y}{1+y}+\frac{3 y}{1+2 y}=0
$$

$y=0$ is not possible, because $x^{0} \neq 10$; thus, we can divide by $y$ and arrive at the following quadratic equation:

$$
y^{2}+5 y+3=0
$$

from which

$$
\begin{gathered}
y_{1,2}=\frac{-5 \pm \sqrt{13}}{2}, \text { and thus } x_{1,2}=10^{\frac{1}{y_{1,2}}} \\
\frac{1}{y_{1,2}}=\frac{2}{-5 \pm \sqrt{13}}=\frac{2(-5 \pm \sqrt{13})}{25-13}=\frac{-5 \pm \sqrt{13}}{6}
\end{gathered}
$$

Thus, the roots obtained are the same as those obtained in the first solution.

Bakos Tamás (Bp. II., Rákóczi g. IV. o. t.)