\section*{Exercise 2 - 041022}

Calculate without using approximate values (without using logarithm tables or slide rule):

\[
y=10-2 \cdot \lg 32-5 \cdot \lg 25
\]
}

The following laws are applied:

(1) \(\lg (a b)=b \cdot \lg a\)

(2) \(\lg (a \cdot b)=\lg a+\lg b\)

(3) \(\lg x=\log _{10} x\)

(4) \(\log _{a} b=c \Leftrightarrow a^{c}=b\) and in particular \(\log _{a} a=1\), because \(a^{1}=a\) and consequently \(\lg 10=1\).

With this, the expression can be equivalently transformed as follows:

\[
\begin{gathered}
y=10-2 \cdot \lg 2^{5}-5 \cdot \lg 5^{2}=10-2 \cdot 5 \cdot \lg 2-5 \cdot 2 \lg 5=10-10 \cdot \lg 2-10 \cdot \lg 5= \\
=10 \cdot(1-\lg 2-\lg 5)=10 \cdot(1-(\lg 2+\lg 5))=10 \cdot(1-(\lg (2 \cdot 5)))=10 \cdot(1-\lg 10)=0
\end{gathered}
\]