Bài tập nâng cao · Bài 19
Bài tập nâng cao Lôgarit · Bài 19
[2D2-0.0-2] ( Chuyên Hùng Vương - Gia Lai - 2018) Tìm bộ ba số nguyên dương $(a\,;\,b;\,c)$ thỏa mãn $\log 1+\log(1+3)+\log(1+3+5)+\,\,.\,.\,.\,\,+\log(1+3+5+\,\,\,.\,.\,.\,\,+19)-2\log 5040=a+b\log 2+c\log 3$
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$\log 1+\log(1+3)+\log(1+3+5)+\,\,.\,.\,.\,\,+\log(1+3+5+\,\,\,.\,.\,.\,\,+19)-2\log 5040=a+b\log 2+c\log 3\Leftrightarrow \log 1+\log 2^{2}+\log 3^{2}+\,\,.\,.\,.\,\,+\log 10^{2}-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow \log\left( 1.2^{2}.3^{2}...10^{2} \right)-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow \log\left( 1.2.3...10 \right)^{2}-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow 2\log\left( 1.2.3...10 \right)-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow 2\left( \log 10!-\log 7! \right)=a+b\log 2+c\log 3\Leftrightarrow 2\log\left( 8.9.10 \right)=a+b\log 2+c\log 3$
$\Leftrightarrow 2+6\log 2+4\log 3=a+b\log 2+c\log 3$.
Vậy $a=2$, $b=6$, $c=4$.
$\log 1+\log(1+3)+\log(1+3+5)+\,\,.\,.\,.\,\,+\log(1+3+5+\,\,\,.\,.\,.\,\,+19)-2\log 5040=a+b\log 2+c\log 3\Leftrightarrow \log 1+\log 2^{2}+\log 3^{2}+\,\,.\,.\,.\,\,+\log 10^{2}-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow \log\left( 1.2^{2}.3^{2}...10^{2} \right)-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow \log\left( 1.2.3...10 \right)^{2}-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow 2\log\left( 1.2.3...10 \right)-2\log 5040=a+b\log 2+c\log 3$
$\Leftrightarrow 2\left( \log 10!-\log 7! \right)=a+b\log 2+c\log 3\Leftrightarrow 2\log\left( 8.9.10 \right)=a+b\log 2+c\log 3$
$\Leftrightarrow 2+6\log 2+4\log 3=a+b\log 2+c\log 3$.
Vậy $a=2$, $b=6$, $c=4$.