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Question Number 74244 by Maclaurin Stickker last updated on 20/Nov/19

$$Findx$$$$x^{{log}_4\sqrt{x}}=x^{{log}_{4}x}-2$$

Answered by MJS last updated on 20/Nov/19

$$t=x^{{\log }_4\sqrt{x}}=x^{\frac{\ln \sqrt{x}}{\ln 4}}=x^{\frac{\ln x}{4\ln 2}}$$$$\ln t=\frac{\ln x}{4\ln 2}\ln x=\frac{{\left(\ln x\right)}^2}{4\ln 2}$$$$\Rightarrow x={\mathrm{e}}^{2\sqrt{\ln 2\ln t}}$$$$x^{{\log }_4\sqrt{x}}=t$$$$x^{{\log }_{4}x}=t^2$$$$t=t^2-2$$$$t^2-t-2=0$$$$\Rightarrow t=-1\vee t=2\mathrm{but}\mathrm{for}x\in \mathbb{R}\mathrm{we}\mathrm{need}t>1$$$$\Rightarrow t=2\Rightarrow x=4$$

Answered by mr W last updated on 20/Nov/19

$$x>0$$$$letu=x^{{\log }_4\sqrt{x}}=x^{\frac{1}{2}{\log }_{4}x}=\sqrt{x^{{\log }_{4}x}}>0$$$$\Rightarrow u=u^2-2$$$$\Rightarrow u^2-u-2=0$$$$\Rightarrow (u-2)(u+1)=0$$$$\Rightarrow u=2,-1$$$$withu=2:$$$$\sqrt{x^{{\log }_{4}x}}=2$$$$x^{{\log }_{4}x}=4$$$$x^{\frac{\ln x}{\ln 4}}=4$$$$\frac{\ln x}{\ln 4}=\frac{\ln 4}{\ln x}$$$${\left(\ln x\right)}^2={\left(\ln 4\right)}^2$$$$\ln x=\pm \ln 4=\ln 4^{\pm 1}$$$$\Rightarrow x=4^{\pm 1}=4or\frac{1}{4}$$

Commented by MJS last updated on 20/Nov/19

$$\mathrm{same}\mathrm{path}\mathrm{we}\mathrm{share}...$$$$\mathrm{but}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{also}\mathrm{has}\mathrm{the}\mathrm{solution}$$$$x=\frac{1}{4}$$$$\mathrm{how}\mathrm{to}\mathrm{get}\mathrm{it}?$$

Commented by mr W last updated on 20/Nov/19

$$twosolutionsindeed.seeabove.$$

Commented by Maclaurin Stickker last updated on 21/Nov/19

$$Sir,Iadded2bythesidesandapplied$$$${log}_{2}inbothsides.Isthiscorrect?$$$$Igotthesameresults$$

Commented by MJS last updated on 21/Nov/19

$${\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{sound}\mathrm{correct}\mathrm{to}\mathrm{me},\mathrm{please}\mathrm{post}\mathrm{it}$$

Commented by Maclaurin Stickker last updated on 21/Nov/19

$$x^{{log}_4\sqrt{x}}+2=x^{{log}_{4}x}$$$${log}_2\left(x^{{log}_4\sqrt{x}}\right)+{log}_{2}2={log}_2\left(x^{{log}_{4}x}\right)$$$$....$$$$\frac{1}{4}{log}_2{x}^2=\frac{1}{2}{log}_2{x}^2-1$$$$....$$$${log}_2{x}^2=4$$$${log}_{2}x=2and{log}_{2}x=-2$$$$x=4orx=\frac{1}{4}$$

Commented by MJS last updated on 21/Nov/19

$$\mathrm{funny}\mathrm{you}\mathrm{get}\mathrm{the}\mathrm{solution}...$$$$\mathrm{but}$$$${\log }_b(p+q)\ne {\log }_{b}p+{\log }_{b}q$$$$b^{{\log }_b(p+q)}=p+a\mathrm{but}{b}^{{\log }_{b}p+{\log }_{b}q}=pq$$$$\mathrm{so}\mathrm{your}{2}^{\mathrm{nd}}\mathrm{line}\mathrm{is}\mathrm{not}\mathrm{equivalent}\mathrm{to}\mathrm{the}\mathrm{first}$$$$\mathrm{line}$$

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