Let $a$ and $b$ be positive numbers, $a \neq 1$. Simplify the following expression:

$$
a^{\lg \left(\frac{1}{b \operatorname{bg} a}\right)}
$$
I. solution: Denoting our expression by $N$ and using the most well-known logarithmic and power identities along with $\lg 1=0$,

$$
N=a^{\lg \left(\frac{1}{b^{\lg a}}\right)}=a^{-\lg \left(b^{\lg a}\right)}=a^{-\lg a \lg b}=\frac{1}{a^{\lg a \lg b}}
$$

We can obtain an even simpler form, with fewer "j" elements, by using the identity

$$
a^{\lg b}=b^{\lg a}
$$

which we derive by interpreting the equality $\lg b \cdot \lg a=\lg a \cdot \lg b$ as the logarithm of powers on both sides. Thus, we have

$$
N=\frac{1}{\left(a^{\lg b}\right)^{\lg a}}=\frac{1}{\left(b^{\lg a}\right)^{\lg a}}=\frac{1}{b^{\lg ^{2} a}}
$$

Nándor Bürger (Bp. XI., József A. g. II. o. t.)

II. solution: Immediately in the first step, by applying the identity (1) (in a reasonable manner):

$$
N=\left(\frac{1}{b^{\lg a}}\right)^{\lg a}=\frac{1}{b^{\lg ^{2} a}}
$$

Ambrus Tihanyi (Bp. V., Apáczai Csere J. gyak. g. II. o. t.)