Bài 2.16 trang 109 SBT giải tích 12

Đề bài

Tìm \(\displaystyle x\), biết:

a) \(\displaystyle{\log _5}x = 2{\log _5}a - 3{\log _5}b\)

b) \(\displaystyle{\log _{\frac{1}{2}}}x = \frac{2}{3}{\log _{\frac{1}{2}}}a - \frac{1}{5}{\log _{\frac{1}{2}}}b\)

Lời giải chi tiết

a) Với \(\displaystyle x,a,b > 0\) thì \(\displaystyle{\log _5}x = 2{\log _5}a - 3{\log _5}b\) \(\displaystyle \Leftrightarrow {\log _5}x = {\log _5}{a^2} - {\log _5}{b^3}\)

\(\displaystyle \Leftrightarrow {\log _5}x = {\log _5}\frac{{{a^2}}}{{{b^3}}}\)\(\displaystyle \Leftrightarrow x = \frac{{{a^2}}}{{{b^3}}}\).

b) Với \(\displaystyle x,a,b > 0\) thì \(\displaystyle{\log _{\frac{1}{2}}}x = \frac{2}{3}{\log _{\frac{1}{2}}}a - \frac{1}{5}{\log _{\frac{1}{2}}}b\)\(\displaystyle \Leftrightarrow {\log _{\frac{1}{2}}}x = {\log _{\frac{1}{2}}}{a^{\frac{2}{3}}} - {\log _{\frac{1}{2}}}{b^{\frac{1}{5}}}\)

\(\displaystyle \Leftrightarrow {\log _{\frac{1}{2}}}x = {\log _{\frac{1}{2}}}\frac{{{a^{\frac{2}{3}}}}}{{{b^{\frac{1}{5}}}}}\) \(\displaystyle \Leftrightarrow x = \frac{{{a^{\frac{2}{3}}}}}{{{b^{\frac{1}{5}}}}}\).

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