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Question Number 213923 by luj last updated on 21/Nov/24

Answered by MATHEMATICSAM last updated on 21/Nov/24

$$x^{32}=2^x$$$$or32\ln \mid x\mid =x\ln 2$$$$\mathrm{For}x>0$$$$or\frac{\ln x}{x}=\frac{\ln 2}{32}$$$$or{x}^{-1}.\ln x=\frac{\ln 2}{32}$$$$or\ln x.e^{-\ln x}=\frac{\ln 2}{32}$$$$or-\ln x.e^{-\ln x}=-\frac{\ln 2}{32}$$$$or-\ln x=W(-\frac{\ln 2}{32})$$$$orx=e^{-W(-\frac{\ln 2}{32})}$$$$\frac{\ln x}{x}=\frac{\ln 2}{32}$$$$or\frac{\ln x}{x}=\frac{{\ln 2}^8}{8\times 32}=\frac{\ln 256}{256}$$$$orx=256$$$$\mathrm{For}x<0$$$$x=e^{-W\left(\frac{\ln 2}{32}\right)}$$

Answered by mr W last updated on 21/Nov/24

$$x^{32}=2^x$$$$x=\pm 2^{\frac{x}{32}}=\pm e^{\frac{x}{32}\ln 2}$$$${xe}^{-\frac{x}{32}\ln 2}=\pm 1$$$$-\frac{x\ln 2}{32}{e}^{-\frac{x}{32}\ln 2}=\pm \frac{\ln 2}{32}$$$$-\frac{x\ln 2}{32}=\mathbb{W}(\pm \frac{\ln 2}{32})$$$$\Rightarrow x=-\frac{32}{\ln 2}\mathbb{W}(\pm \frac{\ln 2}{32})$$$$=\{\begin{array}{l}-\frac{32}{\ln 2}\mathbb{W}\left(\frac{\ln 2}{32}\right)\approx -0.97901693\\ -\frac{32}{\ln 2}\mathbb{W}(-\frac{\ln 2}{32})=\{\begin{array}{l}=256\\ \approx 1.02239294\end{array}\end{array}$$

Answered by kapoorshah last updated on 21/Nov/24

$$x^{\frac{1}{x}}=2^{\frac{1}{32}}$$$$x^{\frac{1}{x}}=2^{\frac{8}{8}\frac{1}{32}}$$$$x^{\frac{1}{x}}=2^{8\left(\frac{1}{8}\frac{1}{32}\right)}$$$$x^{\frac{1}{x}}={256}^{\frac{1}{256}}$$$$x=256$$$$$$$$$$$$$$$$$$

Commented by mr W last updated on 21/Nov/24

$$therearetotallythreerealsolutions.$$

Commented by MATHEMATICSAM last updated on 21/Nov/24

$$\mathrm{yes}$$

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