Пропотенцируйте logx= -log(a+b) + log3 - 1/2logb + 1/3(loga-2/3logb). Created by SerGhost. algebra-ru

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SerGhost @SerGhost

August 2022 1 6 Report

Пропотенцируйте logx= -log(a+b) + log3 - 1/2logb + 1/3(loga-2/3logb)

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NNNLLL54

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$log_{c}x=-log_{c}(a+b)+log_{c}3-\frac{1}{2}\, log_{c}b+\frac{1}{3}\, (log_{c}a-\frac{2}{3}\, log_{c}b)\\\\log_{c}x=log_{c}(a+b)^{-1}+log_{c}3-log_{c}b^{\frac{1}{2}}+log_{c}a^{\frac{1}{3}}-log_{c}b^{\frac{2}{9}}\\\\log_{c}x=log_{c}\frac{3\, \cdot \, a^{\frac{1}{3}}}{(a+b)\, \cdot \, b^{\frac{1}{2}}\, \cdot \, b^{\frac{2}{9}}}\\\\\\x=\dfrac{3\, \cdot \, a^{\frac{1}{3}}}{(a+b)\, \cdot \, b^{\frac{1}{2}}\, \cdot \, b^{\frac{2}{9}}}=\dfrac{3\, \sqrt[3]{a}}{(a+b)\, \cdot \, b^{\frac{13}{18}}}$

$x=\dfrac{3\, \sqrt[3]{a}}{(a+b)\, \cdot \, \sqrt[18]{b^{13}}}\\\\\\P.S.\; \; log_{a}x+log_{a}y=log_{a}(xy)\; \; ,\; \; log_{a}x-log_{a}y=log_{a}\frac{x}{y}\; \; ,\; \; k\cdot log_{a}x=log_{a}x^{k}\; \; .$

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SerGhost August 2021 | 0 Ответы

cherez gipotenuzu pryamokutnogo trikutnika provedena ploshina pid kutom 30 do plosh

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god v yakomu napryamku i z yakoyu shvidkistyu ruhayetsya verhnya tochka shini kolesa avt

cherez gipotenuzu pryamokutnogo trikutnika provedena ploshina pid kutom 30 do plosh

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