Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 792

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\[1)\log_{x^{2}}9 + \log_{\sqrt{x}}4 = 2\]

\[\frac{1}{2}\log_{x}9 + 2\log_{x}4 = 2\]

\[\log_{x}9^{\frac{1}{2}} + \log_{x}4^{2} = 2\]

\[\log_{x}3 + \log_{x}16 = 2\]

\[\log_{x}3 \cdot 16 = 2\]

\[\log_{x}48 = 2\]

\[x^{2} = 48\]

\[x = \pm \sqrt{48}\]

\[Ответ:x = \sqrt{48}.\]

\[2)\log_{x^{2}}16 - \log_{\sqrt{x}}7 = 2\]

\[\frac{1}{2}\log_{x}16 - 2\log_{x}7 = 2\]

\[\log_{x}16^{\frac{1}{2}} - \log_{x}7^{2} = 2\]

\[\log_{x}4 - \log_{x}49 = 2\]

\[\log_{x}\frac{4}{49} = 2\]

\[x^{2} = \frac{4}{49}\]

\[x = \pm \frac{2}{7}.\]

\[Ответ:x = \frac{2}{7}.\]

\[3)\ 2\log_{x}7 - \frac{1}{2}\log_{x^{2}}16 +\]

\[+ \frac{1}{4}\log_{\sqrt{x}}64 = 2\]

\[\log_{x}7^{2} - \frac{1}{4}\log_{x}16 +\]

\[+ \frac{1}{2}\log_{x}64 = 2\]

\[\log_{x}49 - \log_{x}16^{\frac{1}{4}} +\]

\[+ \log_{x}64^{\frac{1}{2}} = 2\]

\[\log_{x}49 - \log_{x}2 + \log_{x}8 = 2\]

\[\log_{x}\frac{49 \cdot 8}{2} = 2\]

\[\log_{x}196 = 2\]

\[x^{2} = 196\]

\[x = \pm 14.\]

\[Ответ:x = 14.\]

\[4)\ \frac{1}{2}\log_{x}7 - \log_{\frac{1}{\sqrt{x}}}{\ 3} -\]

\[- \log_{x^{2}}28 = 1\]

\[\log_{x}7^{\frac{1}{2}} - \log_{x^{- \frac{1}{2}}}3 -\]

\[- \frac{1}{2}\log_{x}28 = 1\]

\[\log_{x}\sqrt{7} + 2\log_{x}3 -\]

\[- \frac{1}{2}\log_{x}28 = 1\]

\[\log_{x}\sqrt{7} + \log_{x}3^{2} - \log_{x}28^{\frac{1}{2}} = 1\]

\[\log_{x}\sqrt{7} + \log_{x}9 - \log_{x}\sqrt{28} = 1\]

\[\log_{x}\frac{\sqrt{7} \cdot 9\ }{\sqrt{28}} = 1\]

\[\log_{x}\frac{9}{2} = 1\]

\[x = \frac{9}{2}.\]

\[Ответ:x = 4,5.\]