Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{\left( {\dfrac{1}{3}} \right)^n} = \dfrac{1}{{81}}\\
 \Leftrightarrow {\left( {\dfrac{1}{3}} \right)^n} = \dfrac{1}{{{3^4}}}\\
 \Leftrightarrow {\left( {\dfrac{1}{3}} \right)^n} = {\left( {\dfrac{1}{3}} \right)^4}\\
 \Leftrightarrow n = 4\\
b,\\
\dfrac{{ - 512}}{{343}} = {\left( {\dfrac{{ - 8}}{7}} \right)^n}\\
 \Leftrightarrow \dfrac{{{{\left( { - 8} \right)}^3}}}{{{7^3}}} = {\left( {\dfrac{{ - 8}}{7}} \right)^n}\\
 \Leftrightarrow {\left( {\dfrac{{ - 8}}{7}} \right)^3} = {\left( {\dfrac{{ - 8}}{7}} \right)^n}\\
 \Leftrightarrow n = 3\\
c,\\
{27^n}:{3^n} = 9\\
 \Leftrightarrow {\left( {27:3} \right)^n} = 9\\
 \Leftrightarrow {9^n} = 9\\
 \Leftrightarrow n = 1\\
d,\\
\dfrac{{81}}{{{{\left( { - 3} \right)}^n}}} =  - 243\\
 \Leftrightarrow 81 = {\left( { - 3} \right)^n}.\left( { - 243} \right)\\
 \Leftrightarrow {\left( { - 3} \right)^4} = {\left( { - 3} \right)^n}.{\left( { - 3} \right)^5}\\
 \Leftrightarrow {\left( { - 3} \right)^4} = {\left( { - 3} \right)^{n + 5}}\\
 \Leftrightarrow 4 = n + 5\\
 \Leftrightarrow n =  - 1\\
e,\\
\dfrac{{25}}{{{5^n}}} = 5\\
 \Leftrightarrow {5^n} = \dfrac{{25}}{5}\\
 \Leftrightarrow {5^n} = 5\\
 \Leftrightarrow n = 1\\
g,\\
\dfrac{1}{2}{.2^n} + {4.2^n} = {9.2^5}\\
 \Leftrightarrow {2^n}.\left( {\dfrac{1}{2} + 4} \right) = {9.2^5}\\
 \Leftrightarrow {2^n}.\dfrac{9}{2} = {9.2^5}\\
 \Leftrightarrow {2^n} = {2^5}.2\\
 \Leftrightarrow {2^n} = {2^6}\\
 \Leftrightarrow n = 6\\
h,\\
{3^{ - 1}}{.4^n} + {3.4^n} = \dfrac{5}{3}{.2^7}\\
 \Leftrightarrow {4^n}.\left( {{3^{ - 1}} + 3} \right) = \dfrac{5}{3}{.2^7}\\
 \Leftrightarrow {4^n}.\left( {\dfrac{1}{3} + 3} \right) = \dfrac{5}{3}{.2^7}\\
 \Leftrightarrow {4^n}.\dfrac{{10}}{3} = \dfrac{5}{3}{.2^7}\\
 \Leftrightarrow {\left( {{2^2}} \right)^n} = \dfrac{1}{2}{.2^7}\\
 \Leftrightarrow {2^{2n}} = {2^6}\\
 \Leftrightarrow 2n = 6\\
 \Leftrightarrow n = 3\\
i,\\
{9^{ - n}}{.27^n} = 243\\
 \Leftrightarrow {\left( {{3^2}} \right)^{ - n}}.{\left( {{3^3}} \right)^n} = {3^5}\\
 \Leftrightarrow {3^{ - 2n}}{.3^{3n}} = {3^5}\\
 \Leftrightarrow {3^{ - 2n + 3n}} = {3^5}\\
 \Leftrightarrow {3^n} = {3^5}\\
 \Leftrightarrow n = 5
\end{array}\)