60 БАЛЛОВ решить с одз log3x+log9x+log27x=11/12. Created by Dgoker2. algebra-ru

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Dgoker2 @Dgoker2

July 2022 1 16 Report

60 БАЛЛОВ решить с одз
log3x+log9x+log27x=11/12

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$log_3x+log_{3^2}x+log_{3^3}x=\frac{11}{12}\; ,\; \; ODZ:\; x>0\\\\log_3x+\frac{1}{2}log_3x+\frac{1}{3}log_3x=\frac{11}{12}\\\\\frac{11}{6}log_3x=\frac{1}{12}\\\\log_3x=\frac{1}{2}\\\\x=3^{\frac{1}{2}}=\sqrt3$

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