2 mũ x = 4 mũ 6 . 16 mũ 3 2 mũ x =4 mũ 4 . 2 mũ 3 2 mũ x = 4 mũ 5 . 16 mũ 2 2 mũ x = 2 mũ 5 . 2 mũ 6 2 mũ x = 16 mũ 5.32 mũ 3 2 mũ x = 32 mũ 5 . 64 m

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
*)\\
{2^x} = {4^6}{.16^3}\\
 \Leftrightarrow {2^x} = {\left( {{2^2}} \right)^6}.{\left( {{2^4}} \right)^3}\\
 \Leftrightarrow {2^x} = {2^{12}}{.2^{12}}\\
 \Leftrightarrow {2^x} = {2^{24}}\\
 \Leftrightarrow x = 24\\
*)\\
{2^x} = {4^4}{.2^3}\\
 \Leftrightarrow {2^x} = {\left( {{2^2}} \right)^4}{.2^3}\\
 \Leftrightarrow {2^x} = {2^8}{.2^3}\\
 \Leftrightarrow {2^x} = {2^{11}}\\
 \Leftrightarrow x = 11\\
*)\\
{2^x} = {4^5}{.16^2}\\
 \Leftrightarrow {2^x} = {\left( {{2^2}} \right)^5}.{\left( {{2^4}} \right)^2}\\
 \Leftrightarrow {2^x} = {2^{10}}{.2^8}\\
 \Leftrightarrow {2^x} = {2^{18}}\\
 \Leftrightarrow x = 18\\
*)\\
{2^x} = {2^5}{.2^6}\\
 \Leftrightarrow {2^x} = {2^{11}}\\
 \Leftrightarrow x = 11\\
*)\\
{2^x} = {16^5}{.32^3}\\
 \Leftrightarrow {2^x} = {\left( {{2^4}} \right)^5}.{\left( {{2^5}} \right)^3}\\
 \Leftrightarrow {2^x} = {2^{20}}{.2^{15}}\\
 \Leftrightarrow {2^x} = {2^{35}}\\
 \Leftrightarrow x = 35\\
*)\\
{2^x} = {32^5}{.64^6}\\
 \Leftrightarrow {2^x} = {\left( {{2^5}} \right)^5}.{\left( {{2^6}} \right)^6}\\
 \Leftrightarrow {2^x} = {2^{25}}{.2^{36}}\\
 \Leftrightarrow {2^x} = {2^{61}}\\
 \Leftrightarrow x = 61\\
*)\\
{2^x} = {4^3}{.8^4}{.16^5}\\
 \Leftrightarrow {2^x} = {\left( {{2^2}} \right)^3}.{\left( {{2^3}} \right)^4}.{\left( {{2^4}} \right)^5}\\
 \Leftrightarrow {2^x} = {2^6}{.2^{12}}{.2^{20}}\\
 \Leftrightarrow {2^x} = {2^{38}}\\
 \Leftrightarrow x = 38\\
*)\\
{3^x} = {9^{ – 6}}{.27^{ – 5}}{.81^8}\\
 \Leftrightarrow {3^x} = {\left( {{3^2}} \right)^{ – 6}}.{\left( {{3^3}} \right)^{ – 5}}.{\left( {{3^4}} \right)^8}\\
 \Leftrightarrow {3^x} = {3^{ – 12}}{.3^{ – 15}}{.3^{32}}\\
 \Leftrightarrow {3^x} = {3^5}\\
 \Leftrightarrow x = 5\\
*)\\
{2^x} = {8^3}{.8^{ – 10}}{.8^3}\\
 \Leftrightarrow {2^x} = {8^{ – 4}}\\
 \Leftrightarrow {2^x} = {\left( {{2^3}} \right)^{ – 4}}\\
 \Leftrightarrow {2^x} = {2^{ – 12}}\\
 \Leftrightarrow x =  – 12
\end{array}\)