Question

Tìm số nguyên x > 0 thỏa mãn \(3.3^2.3^3.3^4....3^x=3^{190}\)

Answer 1

\(3\cdot3^2\cdot3^3\cdot3^4\cdot....\cdot3^x=3^{190}\)

\(\Leftrightarrow3^{1+2+3+...+x}=3^{190}\)

\(\Leftrightarrow1+2+3+...+x=190\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2}=190\Leftrightarrow x\left(x+1\right)=380\)

\(\Leftrightarrow x^2+x-380=0\)

\(\Leftrightarrow x^2-19x+20x-380=0\)

\(\Leftrightarrow x\left(x-19\right)+20\left(x-19\right)=0\)

\(\Leftrightarrow\left(x-19\right)\left(x+20\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-19=0\\x+20=0\end{matrix}\right.\)\(\Leftrightarrow x=19\left(x>0\right)\)

Answer 2

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