Question

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log3x^3+log2x^2=2lg6/lg2+1

Answer 1

$log_3x^3+log_2x^2=2\cdot \frac{lg6}{lg2}+1\; \; ,\; \; \; ODZ:\; x>0\; ,\\\\\\\frac{lgx^3}{lg3}+\frac{lgx^2}{lg2}=2\cdot \frac{lg6}{lg2}+1\\\\\frac{3\cdot lg2\cdot lgx+2\cdot lg3\cdot lgx}{lg3\, \cdot \, lg2}=\frac{2\cdot lg6+lg2}{lg2}\\\\\frac{lgx\cdot (3\cdot lg2+2\cdot lg3)}{lg3\, \cdot \, lg2}=\frac{2\cdot (lg2+lg3)+lg2}{lg2}\\\\lgx=\frac{3\cdot lg2+2\cdot lg3}{lg2}\cdot \frac{lg3\, \cdot \, lg2}{3\cdot lg2+2\cdot lg3}\\\\lgx=lg3\\\\x=3\\\\Otvet:\; \; x=3\; .$