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Question Number 88068 by student work last updated on 08/Apr/20

Commented by john santu last updated on 08/Apr/20

$$doyoumean\ln x^2.{2\log }_{2x}\left(x\right)={\log }_{4x}\left(2\right)??$$

Commented by $@ty@m123 last updated on 08/Apr/20

$$Whatisthebasein\log x^2?$$

Commented by john santu last updated on 08/Apr/20

$$iguessyourquestionis$$$${\log }_{x^2}\left(2\right).{\log }_{2x}\left(2\right)={\log }_{4x}\left(2\right)$$

Answered by MJS last updated on 08/Apr/20

$$\mathrm{if}\mathrm{it}\mathrm{is}$$$$\ln x^2\times {2\log }_{2x}2={\log }_{4x}2$$$$\mathrm{then}:$$$$2\ln x\times 2\frac{\ln 2}{\ln 2x}=\frac{\ln 2}{\ln 4x}$$$$4\frac{\ln x}{\ln 2x}=\frac{1}{\ln 4x}$$$$4\ln x\ln 4x=\ln 2x$$$$4\ln x(\ln x+2\ln 2)=\ln x+\ln 2$$$$\mathrm{let}t=\ln x$$$$4t(t+2\ln 2)=t+\ln 2$$$$t^2-(\frac{1}{4}-2\ln 2)t-\frac{\ln 2}{4}=0$$$$\Rightarrow$$$$t=\frac{1}{8}-\ln 2\pm \sqrt{\frac{1}{64}+{\left(\ln 2\right)}^2}$$$$\Rightarrow$$$$x=\frac{1}{2}{\mathrm{e}}^{\frac{1}{8}\pm \sqrt{\frac{1}{64}+{\left(\ln 2\right)}^2}}$$$$x_1\approx .280137$$$$x_2\approx 1.14589$$

Commented by Zainal Arifin last updated on 13/Apr/20

$$\mathrm{thank}\mathrm{sir}$$

Answered by MJS last updated on 08/Apr/20

$$\mathrm{if}\mathrm{it}\mathrm{is}$$$${\log }_{x^2}2\times {\log }_{2x}2={\log }_{4x}2$$$$\mathrm{then}$$$$\frac{\ln 2}{\ln x^2}\times \frac{\ln 2}{\ln 2x}=\frac{\ln 2}{\ln 4x}$$$$\frac{\ln 2}{2\ln x(\ln x+\ln 2)}=\frac{1}{\ln 4x}$$$$...\mathrm{let}t=\ln x$$$$t^2+\frac{\ln 2}{2}t-{\left(\ln 2\right)}^2=0$$$$\Rightarrow t=-\frac{1\pm \sqrt{17}}{4}\ln 2$$$$\Rightarrow x=\frac{1}{2}{2}^{\frac{3\pm \sqrt{17}}{4}}$$$$x_1\approx .411574$$$$x_2\approx 1.71806$$

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