Giải bài 2.16 trang 109 - SBT Giải tích lớp 12
Tìm x, biết:
\(a)\,\log_5x=2\log_5a-3\log_5b\)
\(b)\,\log_{\frac 1 2}x=\dfrac 2 3\log_{\frac 1 2 }a-\dfrac 1 5 \log_{\frac 1 2}b\)
a)
\(\begin{align} & {{\log }_{5}}x=2{{\log }_{5}}a-3{{\log }_{5}}b \\ & \Leftrightarrow {{\log }_{5}}x={{\log }_{5}}{{a}^{2}}-{{\log }_{5}}{{b}^{3}} \\ & \Leftrightarrow {{\log }_{5}}x={{\log }_{5}}\dfrac{{{a}^{2}}}{{{b}^{3}}} \\ & \Leftrightarrow x=\dfrac{{{a}^{2}}}{{{b}^{3}}} \\ \end{align} \)
b)
\(\begin{align} & {{\log }_{\frac{1}{2}}}x=\dfrac{2}{3}{{\log }_{\frac{1}{2}}}a-\dfrac{1}{5}{{\log }_{\frac{1}{2}}}b \\ & \Leftrightarrow {{\log }_{\frac{1}{2}}}x={{\log }_{\frac{1}{2}}}{{a}^{\frac{2}{3}}}-{{\log }_{\frac{1}{2}}}{{b}^{\frac{1}{5}}} \\ & \Leftrightarrow {{\log }_{\frac{1}{2}}}x={{\log }_{\frac{1}{2}}}\dfrac{{{a}^{\frac{2}{3}}}}{{{b}^{\frac{1}{5}}}} \\ & \Leftrightarrow x=\dfrac{{{a}^{\frac{2}{3}}}}{{{b}^{\frac{1}{5}}}}=\dfrac{\sqrt[3]{{{a}^{2}}}}{\sqrt[5]{b}} \\ \end{align} \)