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Question Number 114597 by O Predador last updated on 19/Sep/20

$$$$$$\mathbf{What}\mathbf{is}\mathbf{the}\mathbf{number}\mathbf{value}\mathbf{of}\mathbf{f}\left[\sqrt[3]{\mathbf{log}\left(\frac{\sqrt{\mathbf{x}}-1}{\mathbf{x}-1}\right)}\right]=\sqrt{\sqrt{\mathbf{x}}+\mathbf{x}}\mathbf{for}\mathbf{f}(-1)?$$$$$$$$\mathbf{a})0,1$$$$\mathbf{b})27$$$$\mathbf{c})81$$$$\mathbf{d})10$$$$\mathbf{e})12$$$$$$

Answered by Olaf last updated on 19/Sep/20

$$\sqrt[3]{\log \left(\frac{\sqrt{x}-1}{x-1}\right)}=-1$$$$\log \left(\frac{\sqrt{x}-1}{x-1}\right)=-1$$$$\frac{\sqrt{x}-1}{x-1}={10}^{-1}=\frac{1}{10}$$$$\sqrt{x}-1=\frac{1}{10}(x-1)$$$$x-10\sqrt{x}+9=0$$$$(\sqrt{x}-1)(\sqrt{x}-9)=0$$$$\sqrt{x}=1:\mathrm{impossible}$$$$\mathrm{otherwise}\frac{\sqrt{x}-1}{x-1}\mathrm{is}\mathrm{not}\mathrm{defined}$$$$\sqrt{x}=9\Rightarrow \sqrt{\sqrt{x}+x}=\sqrt{90}=3\sqrt{10}$$$$\mathrm{Sorry}\mathrm{but}\mathrm{I}\mathrm{find}3\sqrt{10}$$$$\mathrm{I}\mathrm{supposed}\log \mathrm{is}{\log }_{10}\mathrm{and}\mathrm{not}\ln$$

Commented by O Predador last updated on 20/Sep/20

$$\mathrm{Thank}\mathrm{you}!$$

Answered by floor(10²Eta[1]) last updated on 20/Sep/20

$$\mathrm{f}\left(\sqrt[3]{\log \left(\frac{\sqrt{\mathrm{x}}-1}{\mathrm{x}-1}\right)}\right)=\sqrt{\sqrt{\mathrm{x}}+\mathrm{x}}$$$$\sqrt[3]{\log \left(\frac{\sqrt{\mathrm{x}}-1}{\mathrm{x}-1}\right)}=-1\Rightarrow \log \left(\frac{\sqrt{\mathrm{x}}-1}{\mathrm{x}-1}\right)=-1$$$$\frac{\sqrt{\mathrm{x}}-1}{\mathrm{x}-1}={10}^{-1}=\frac{1}{10}\Rightarrow 10\sqrt{\mathrm{x}}-10=\mathrm{x}-1$$$$10\sqrt{\mathrm{x}}=\mathrm{x}+9$$$$100\mathrm{x}={\mathrm{x}}^2+18\mathrm{x}+81$$$${\mathrm{x}}^2-82\mathrm{x}+81=0$$$$\mathrm{x}=\frac{82\pm 80}{2}\Rightarrow \mathrm{x}=1\mathrm{or}81\Rightarrow \mathrm{x}=81(\mathrm{because}\mathrm{x}\ne 1)$$$$\mathrm{x}=81\Rightarrow$$$$\mathrm{f}(-1)=\sqrt{\sqrt{81}+81}=\sqrt{90}$$$$\mathrm{now}\mathrm{if}\log \mathrm{is}\ln \mathrm{the}\mathrm{logic}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{and}$$$$\mathrm{the}\mathrm{answer}\mathrm{is}\sqrt{{\mathrm{e}}^2-\mathrm{e}}$$$$$$

Answered by 1549442205PVT last updated on 20/Sep/20

$$\mathrm{We}\mathrm{need}\mathrm{find}\mathrm{x}>0,\mathrm{x}\ne 1\mathrm{such}\mathrm{that}$$$$\log \frac{\sqrt{\mathrm{x}}-1}{\mathrm{x}-1}=-1\iff \log \frac{\sqrt{\mathrm{x}}-1}{(\sqrt{\mathrm{x}}-1)(\sqrt{\mathrm{x}}+1)}=-1$$$$\iff \log \frac{1}{\sqrt{\mathrm{x}}+1}=-1\iff -\log (\sqrt{\mathrm{x}}+1)=-1$$$$\bullet \mathrm{If}\mathrm{logx}={\log }_{10}\mathrm{x}\mathrm{then}$$$$\iff \log (\sqrt{\mathrm{x}}+1)=1\Rightarrow \sqrt{\mathrm{x}}+1=10$$$$\iff \sqrt{\mathrm{x}}=9\iff \mathrm{x}=81$$$$\Rightarrow \mathrm{f}(-1)=\sqrt{9+81}=\sqrt{90}$$$$\bullet \mathrm{If}\mathrm{logx}=\mathrm{lnx}\mathrm{then}\sqrt{\mathrm{x}}+1=\mathrm{e}\Rightarrow \sqrt{\mathrm{x}}=\mathrm{e}-1$$$$\Rightarrow \mathrm{x}={\mathrm{e}}^2-2\mathrm{e}+1\Rightarrow \mathrm{f}(-1)=\sqrt{{\mathrm{e}}^2-\mathrm{e}}$$$$\mathbf{Available}\mathbf{answer}\mathbf{is}\mathbf{false}!$$

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