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SchoolFrankinshtein @SchoolFrankinshtein

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Решить уравнение: 1) log(у основания 3)x = 4; 2) log(у основания x)8 = 1/3.

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$log_3x=4\; \; ,\; \; x=3^4\; ,\; \; x=81\; \; \; (x>0)\\\\\\log_{x}8=\frac{1}{3}\; \; ,\; \; \frac{1}{log_8x}=\frac{1}{3}\; \; ,\; \; log_8x=3\; ,\; \; x=8^3\; ,\; \; x=512\; \; \; (x>0\; ,\; x\ne 1)$

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