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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