Câu hỏi:
Tìm số cặp số nguyên \(\left( {a;b} \right)\) thỏa mãn \({\log _a}b + 6{\log _b}a = 5\), biết: \(2 \le a \le 2020;2 \le b \le 2021\).
A. \(54\)
B. \(55\).
C. \(2021\).
D. \(4041\).
LỜI GIẢI CHI TIẾT
\({\log _a}b + 6{\log _b}a = 5 \Leftrightarrow {\log _a}b + \frac{6}{{{{\log }_a}b}} = 5 \Leftrightarrow {\left( {{{\log }_a}b} \right)^2} – 5{\log _a}b + 6 = 0\)
\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{{{\log }_a}b = 3}\\{{{\log }_a}b = 2}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{b = {a^3}}\\{b = {a^2}}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{a = \sqrt[3]{b}}\\{a = \sqrt b }\end{array}} \right.\)
TH1: \(\left\{ {\begin{array}{*{20}{c}}{2 \le a \le 2020}\\{2 \le b \le 2021}\\{a = \sqrt[3]{b}}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{2 \le a \le 2020}\\{\sqrt[3]{2} \le \sqrt[3]{b} \le \sqrt[3]{{2021}}}\\{a = \sqrt[3]{b}}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}}{2 \le a \le 2020}\\{\sqrt[3]{2} \le a \le \sqrt[3]{{2021}}}\\{a = \sqrt[3]{b}}\end{array}} \right.} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{2 \le a \le 12}\\{b = {a^3}}\end{array}} \right.\)
Do đó có \(11\) cặp số \(\left( {a;b} \right)\) thỏa yêu cầu.
TH2: \(\left\{ {\begin{array}{*{20}{c}}{2 \le a \le 2020}\\{2 \le b \le 2021}\\{a = \sqrt b }\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{2 \le a \le 2020}\\{\sqrt 2 \le \sqrt b \le \sqrt {2021} }\\{a = \sqrt b }\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}}{2 \le a \le 2020}\\{\sqrt 2 \le a \le \sqrt {2021} }\\{a = \sqrt b }\end{array}} \right.} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{2 \le a \le 44}\\{b = {a^2}}\end{array}} \right.\)
Do đó có \(43\) cặp số \(\left( {a;b} \right)\) thỏa yêu cầu.
Vậy: có \(54\) cặp số \(\left( {a;b} \right)\) thỏa yêu cầu.
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