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--- jancraats:hoofdstuk_18 [2012/05/15 19:23] – randy
+++ jancraats:hoofdstuk_18 [2022/09/01 11:44] (current) – external edit 127.0.0.1
@@ Line 3: / Line 3: @@
 $$
 \begin{split}
-^{1 + x} \cdot 3^x = \\
+&2^{1 + x} &\cdot 3^x  \\
-^1 \cdot 2^x \cdot 3^x \\
+= & 2^1 \cdot 2^x &\cdot 3^x  \\
-^1 \cdot 6^x \\
+= & 2^1 &\cdot 6^x  \\
-\cdot 6^x
+= & 2 &\cdot 6^x
+\end{split}
+$$
+====== 18.1e ======
+$$
+\begin{split}
+& 4^{2x} : 2^x \\
+= &(4^2)^x : 2^x \\
+= &16^x : 2^x \\
+= & \left( \frac{16}{2} \right) ^x \\
+= & 8^x
 \end{split}
 $$