Giải thích các bước giải:
\[\begin{array}{l}
c,\\
\frac{{16}}{{{2^n}}} = 2\\
\Leftrightarrow {2^n} = \frac{{16}}{2}\\
\Leftrightarrow {2^n} = 8\\
\Leftrightarrow {2^n} = {2^3}\\
\Leftrightarrow n = 3\\
d,\\
\frac{{{{\left( { - 3} \right)}^n}}}{{81}} = - 27\\
\Leftrightarrow \frac{{{{\left( { - 3} \right)}^n}}}{{{{\left( { - 3} \right)}^4}}} = {\left( { - 3} \right)^3}\\
\Leftrightarrow {\left( { - 3} \right)^n} = {\left( { - 3} \right)^7}\\
\Leftrightarrow n = 7\\
e,\\
{8^n}:{2^n} = 4\\
\Leftrightarrow {\left( {{2^3}} \right)^n}:{2^n} = {2^2}\\
\Leftrightarrow {2^{3n - n}} = {2^2}\\
\Leftrightarrow {2^{2n}} = {2^2}\\
\Rightarrow n = 1\\
h,\\
{3^{ - 2}}{.3^4}{.3^n} = {3^7}\\
\Leftrightarrow - 2 + 4 + n = 7\\
\Leftrightarrow n = 5\\
i,\\
{2^{ - 1}}{.2^n} + {4.2^n} = {9.2^5}\\
\Leftrightarrow {2^{n - 1}} + {4.2^n} = {9.2^5}\\
\Leftrightarrow {2^{n - 1}}\left( {1 + 4.2} \right) = {9.2^5}\\
\Leftrightarrow n - 1 = 5\\
\Leftrightarrow n = 6
\end{array}\]
