Lösung 1.3:6f
Aus Online Mathematik Brückenkurs 1
We can factorize the exponents \displaystyle 40 and \displaystyle 56 as
\displaystyle \begin{align}
& 40=4\centerdot 10=2\centerdot 2\centerdot 2\centerdot 5=2^{3}\centerdot 5 \\
& \\
& 56=7\centerdot 8=7\centerdot 2\centerdot 4=7\centerdot 2\centerdot 2\centerdot 2=2^{3}\centerdot 7 \\
\end{align}
and we then see that they have
\displaystyle 2^{3}=8
as a common factor. We can take this factor out as an "outer" exponent:
\displaystyle \begin{align}
& 3^{40}=3^{5\centerdot 8}=\left( 3^{5} \right)^{8}=\left( 3\centerdot 3\centerdot 3\centerdot 3\centerdot 3 \right)^{8}=243^{8} \\
& \\
& 2^{56}=2^{7\centerdot 8}=\left( 2^{7} \right)^{8}=\left( 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2 \right)^{8}=128^{8} \\
\end{align}
This shows that
\displaystyle 3^{40}=243^{8}
is bigger than
\displaystyle 2^{56}=128^{8}