For which values of the parameter $m$ will the equation

$$
0.5^{x^{2}-m x+0.5 m-1.5}=(\sqrt{8})^{m-1}
$$

have real and distinct roots?
Both sides of the equation can easily be written as powers of the common base $2$, since $0.5=2^{-1}$ and $\sqrt{8}=2^{3/2}=2^{1.5}$:

$$
2^{-x^{2}+m x-0.5 m+1.5}=2^{1.5 m-1.5}
$$

This can only hold if the exponents are also equal:

$$
\begin{gathered}
-x^{2}+m x-0.5 m+1.5=1.5 m-1.5 \\
x^{2}-m x+2 m-3=0
\end{gathered}
$$

The roots of the resulting quadratic equation are real and distinct if and only if the discriminant of the equation is positive, that is,

$$
D=m^{2}-8 m+12>0
$$

The value of $D$ becomes zero at $m_{1}=2$ and $m_{2}=6$, so the quadratic polynomial in $m$ can be factored as (2):

$$
D=(m-2)(m-6)>0
$$

This product is positive if and only if its factors have the same sign. Both are negative if the larger $m-2$ factor is negative, and both are positive if the smaller $m-6$ factor is positive, that is, when

$$
\begin{array}{ll}
m-2<0 \text {; from these } \\
m-6>0 \text {; from these } \\
m<2 \text { and } \\
m>6 .
\end{array}
$$

Thus, we can state our result as follows: The roots of (1) are real and distinct for all $m$ except those satisfying $2 \leq m \leq 6$.

József Kertész (Miskolc-Diósgyőr, Kilián Gy. g. III. o. t.)

Remark. Solutions that are not essentially different from the above are those that started by writing that the logarithm to an arbitrary base $b$ (where $b>0$ and $b \neq 1$) of both sides of (1) is also equal.