Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\left( {3x - 21} \right){.6^{11}} = {2.6^{12}}\\
\Leftrightarrow 3x - 21 = {2.6^{12}}:{6^{11}}\\
\Leftrightarrow 3x - 21 = {2.6^{12 - 11}}\\
\Leftrightarrow 3x - 21 = {2.6^1}\\
\Leftrightarrow 3x - 21 = 12\\
\Leftrightarrow 3x = 12 + 21\\
\Leftrightarrow 3x = 33\\
\Leftrightarrow x = 33:3\\
\Leftrightarrow x = 11\\
b,\\
{5.2^x} = 320\\
\Leftrightarrow {2^x} = 320:5\\
\Leftrightarrow {2^x} = 64\\
\Leftrightarrow {2^x} = {2^6}\\
\Leftrightarrow x = 6\\
c,\\
{4.3^{x - 1}} = 36\\
\Leftrightarrow {3^{x - 1}} = 36:4\\
\Leftrightarrow {3^{x - 1}} = 9\\
\Leftrightarrow {3^{x - 1}} = {3^2}\\
\Leftrightarrow x - 1 = 2\\
\Leftrightarrow x = 2 + 1\\
\Leftrightarrow x = 3\\
d,\\
{2^x} + {5.2^x} = 48\\
\Leftrightarrow {2^x}.\left( {1 + 5} \right) = 48\\
\Leftrightarrow {2^x}.6 = 48\\
\Leftrightarrow {2^x} = 48:6\\
\Leftrightarrow {2^x} = 8\\
\Leftrightarrow {2^x} = {2^3}\\
\Leftrightarrow x = 3\\
e,\\
{3^{x + 1}} + {5.3^x} = 72\\
\Leftrightarrow {3^x}{.3^1} + {5.3^x} = 72\\
\Leftrightarrow {3^x}.\left( {3 + 5} \right) = 72\\
\Leftrightarrow {3^x}.8 = 72\\
\Leftrightarrow {3^x} = 72:8\\
\Leftrightarrow {3^x} = 9\\
\Leftrightarrow {3^x} = {3^2}\\
\Leftrightarrow x = 2\\
f,\\
625:{19^x} = {25^2}\\
\Leftrightarrow {19^x} = 625:{25^2}\\
\Leftrightarrow {19^x} = 1\\
\Leftrightarrow x = 0\\
i,\\
{\left( {x - 5} \right)^{50}} = {\left( {x - 5} \right)^{18}}\\
\Rightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x - 5 = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 5\\
x = 6
\end{array} \right.\\
k,\\
{\left( {{x^2} - 7} \right)^2} = 4\\
\Leftrightarrow {\left( {{x^2} - 7} \right)^2} = {2^2}\\
\Leftrightarrow {x^2} - 7 = 2\\
\Leftrightarrow {x^2} = 2 + 7\\
\Leftrightarrow {x^2} = 9\\
\Leftrightarrow x = 3\\
l,\\
{\left( {{x^3} - 5} \right)^2} = 9\\
\Leftrightarrow {\left( {{x^3} - 5} \right)^2} = {3^2}\\
\Leftrightarrow {x^3} - 5 = 3\\
\Leftrightarrow {x^3} = 3 + 5\\
\Leftrightarrow {x^3} = 8\\
\Leftrightarrow {x^3} = {2^3}\\
\Leftrightarrow x = 2
\end{array}\)
\(\begin{array}{l}
m,\\
{\left( {{x^2} - 6} \right)^4} = 81\\
\Leftrightarrow {\left( {{x^2} - 6} \right)^4} = {3^4}\\
\Leftrightarrow {x^2} - 6 = 3\\
\Leftrightarrow {x^2} = 3 + 6\\
\Leftrightarrow {x^2} = 9\\
\Leftrightarrow x = 3\\
n,\\
16 \le {2^x} \le 128\\
\Leftrightarrow {2^4} \le {2^x} \le {2^7}\\
\Leftrightarrow 4 \le x \le 7\\
\Rightarrow x \in \left\{ {4;5;6;7} \right\}
\end{array}\)
