Google
Câu hỏi: Có tất cả bao nhiêu số dương a thỏa mãn đẳng thức sau:
${{\log }_{2}}a+{{\log }_{3}}a+{{\log }_{5}}a={{\log }_{2}}a.{{\log }_{3}}a.{{\log }_{5}}a?$
A. 1.
B. 0.
C. 3.
D. 2.
${{\log }_{2}}a+{{\log }_{3}}a+{{\log }_{5}}a={{\log }_{2}}a.{{\log }_{3}}a.{{\log }_{5}}a?$
A. 1.
B. 0.
C. 3.
D. 2.
Điều kiện: $a>0$
Ta có: ${{\log }_{2}}a+{{\log }_{3}}a+{{\log }_{5}}a={{\log }_{2}}a.{{\log }_{3}}a.{{\log }_{5}}a$
$\Leftrightarrow {{\log }_{2}}a+{{\log }_{3}}2.{{\log }_{2}}a+{{\log }_{5}}2.{{\log }_{2}}a={{\log }_{2}}a.{{\log }_{3}}2.{{\log }_{2}}a.{{\log }_{5}}2.{{\log }_{2}}a$
$\Leftrightarrow {{\log }_{2}}a\left( 1+{{\log }_{3}}2+{{\log }_{5}}2 \right)=\log _{2}^{3}a.{{\log }_{3}}2.{{\log }_{5}}2$
$\Leftrightarrow {{\log }_{2}}a\left( \log _{2}^{2}a.{{\log }_{3}}2.{{\log }_{5}}2-1-{{\log }_{3}}2-{{\log }_{5}}2 \right)=0$
$\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{2}}a=0 \\
& \log _{2}^{2}.{{\log }_{3}}2.{{\log }_{5}}2-1-{{\log }_{3}}2-{{\log }_{5}}2=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& a=1 \\
& \log _{2}^{2}a=\dfrac{1+{{\log }_{3}}2+{{\log }_{5}}2}{{{\log }_{3}}2.{{\log }_{5}}2} \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& a=1 \\
& {{\log }_{2}}a=\sqrt{\dfrac{1+{{\log }_{3}}2+{{\log }_{5}}2}{{{\log }_{3}}2.{{\log }_{5}}2}}={{t}_{1}} \\
& {{\log }_{2}}a=-\sqrt{\dfrac{1+{{\log }_{3}}2+{{\log }_{5}}2}{{{\log }_{3}}2.{{\log }_{5}}2}}={{t}_{2}} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& a=1 \\
& a={{2}^{{{t}_{1}}}}>0 \\
& a={{2}^{{{t}_{2}}}}>0 \\
\end{aligned} \right.$
Vậy phương trình đã cho có 3 nghiệm $a>0.$
Ta có: ${{\log }_{2}}a+{{\log }_{3}}a+{{\log }_{5}}a={{\log }_{2}}a.{{\log }_{3}}a.{{\log }_{5}}a$
$\Leftrightarrow {{\log }_{2}}a+{{\log }_{3}}2.{{\log }_{2}}a+{{\log }_{5}}2.{{\log }_{2}}a={{\log }_{2}}a.{{\log }_{3}}2.{{\log }_{2}}a.{{\log }_{5}}2.{{\log }_{2}}a$
$\Leftrightarrow {{\log }_{2}}a\left( 1+{{\log }_{3}}2+{{\log }_{5}}2 \right)=\log _{2}^{3}a.{{\log }_{3}}2.{{\log }_{5}}2$
$\Leftrightarrow {{\log }_{2}}a\left( \log _{2}^{2}a.{{\log }_{3}}2.{{\log }_{5}}2-1-{{\log }_{3}}2-{{\log }_{5}}2 \right)=0$
$\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{2}}a=0 \\
& \log _{2}^{2}.{{\log }_{3}}2.{{\log }_{5}}2-1-{{\log }_{3}}2-{{\log }_{5}}2=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& a=1 \\
& \log _{2}^{2}a=\dfrac{1+{{\log }_{3}}2+{{\log }_{5}}2}{{{\log }_{3}}2.{{\log }_{5}}2} \\
\end{aligned} \right.$
$\Leftrightarrow \left[ \begin{aligned}
& a=1 \\
& {{\log }_{2}}a=\sqrt{\dfrac{1+{{\log }_{3}}2+{{\log }_{5}}2}{{{\log }_{3}}2.{{\log }_{5}}2}}={{t}_{1}} \\
& {{\log }_{2}}a=-\sqrt{\dfrac{1+{{\log }_{3}}2+{{\log }_{5}}2}{{{\log }_{3}}2.{{\log }_{5}}2}}={{t}_{2}} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& a=1 \\
& a={{2}^{{{t}_{1}}}}>0 \\
& a={{2}^{{{t}_{2}}}}>0 \\
\end{aligned} \right.$
Vậy phương trình đã cho có 3 nghiệm $a>0.$
| Công thức đổi cơ số: ${{\log }_{a}}c={{\log }_{a}}b.{{\log }_{b}}c$ |
Đáp án C.