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Câu hỏi: Cho số thực x thỏa mãn : $\log x=\dfrac{1}{2}\log 3a-2\log b+3\log \sqrt{c}$ (a, b, c là các số thực dương). Hãy biểu diễn x theo a, b, c.
A. $x=\dfrac{{{c}^{3}}\sqrt{3a}}{{{b}^{2}}}$.
B. $x=\dfrac{\sqrt{3a}}{{{b}^{2}}{{c}^{3}}}$.
C. $x=\dfrac{\sqrt{3ac}}{{{b}^{2}}}$.
D. $x=\dfrac{\sqrt{3a{{c}^{3}}}}{{{b}^{2}}}$.
Ta có : $\log x=\dfrac{1}{2}\log 3a-2\log b+3\log \sqrt{c}$ $\Leftrightarrow \log x=\log \sqrt{3a}-\log {{b}^{2}}+\log \sqrt{{{c}^{3}}}$.
$\Leftrightarrow \log x=\log \dfrac{\sqrt{3a}.\sqrt{{{c}^{3}}}}{{{b}^{2}}}\Leftrightarrow x=\dfrac{\sqrt{3a{{c}^{3}}}}{{{b}^{2}}}.$
A. $x=\dfrac{{{c}^{3}}\sqrt{3a}}{{{b}^{2}}}$.
B. $x=\dfrac{\sqrt{3a}}{{{b}^{2}}{{c}^{3}}}$.
C. $x=\dfrac{\sqrt{3ac}}{{{b}^{2}}}$.
D. $x=\dfrac{\sqrt{3a{{c}^{3}}}}{{{b}^{2}}}$.
Ta có : $\log x=\dfrac{1}{2}\log 3a-2\log b+3\log \sqrt{c}$ $\Leftrightarrow \log x=\log \sqrt{3a}-\log {{b}^{2}}+\log \sqrt{{{c}^{3}}}$.
$\Leftrightarrow \log x=\log \dfrac{\sqrt{3a}.\sqrt{{{c}^{3}}}}{{{b}^{2}}}\Leftrightarrow x=\dfrac{\sqrt{3a{{c}^{3}}}}{{{b}^{2}}}.$
Đáp án D.