[tex] \bf 32^{2c}=8^{c+7}~~
\begin{cases}
32=2\cdot 2\cdot 2\cdot 2\cdot 2\\
\qquad 2^5\\
8=2\cdot 2\cdot 2\\
\qquad 2^3
\end{cases}\implies (2^5)^{2c}=(2^3)^{c+7}
\\\\\\
2^{5(2c)}=2^{3(c+7)}\implies \stackrel{\textit{if the bases are the same, then so are the exponents}}{5(2c)=3(c+7)\implies 10c=3c+21}
\\\\\\
7c=21\implies c=\cfrac{21}{7}\implies c=3 [/tex]
