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Question Number 112808 by Aina Samuel Temidayo last updated on 09/Sep/20

Answered by 1549442205PVT last updated on 10/Sep/20

$$\mathrm{We}\mathrm{have}{\log }_2(10+2\mathrm{x})={\log }_{10}(\mathrm{x}-4)+2$$$$\iff [{\log }_2(10+2\mathrm{x})-2]-{\log }_{10}(\mathrm{x}-4)=0\left(1\right)$$$$\mathrm{We}\mathrm{need}\mathrm{the}\mathrm{condition}\mathrm{x}>4.\mathrm{Then}$$$${\log }_2(10+2\mathrm{x})-2={\log }_4{(10+2\mathrm{x})}^2-2$$$$⩾{\log }_4\left(80\mathrm{x}\right)-2={\log }_4\left(5\mathrm{x}\right)\left(2\right)$$$$(\mathrm{since}{(\mathrm{a}+\mathrm{b})}^2⩾4\mathrm{ab}\Rightarrow {(10+2\mathrm{x})}^2⩾80\mathrm{x})$$$$\mathrm{Now}\mathrm{we}\mathrm{prove}{\log }_4\left(5\mathrm{x}\right)>{\log }_{10}(\mathrm{x}-4)$$$$\mathrm{Indeed},\mathrm{we}\mathrm{have}{\log }_4\left(5\mathrm{x}\right)>{\log }_{10}\left(5\mathrm{x}\right)\left(3\right)$$$$\mathrm{but}\mathrm{since}\mathrm{x}>4,\mathrm{so}5\mathrm{x}>\mathrm{x}-4.\mathrm{This}$$$$\mathrm{implies}\mathrm{that}{\log }_{10}\left(5\mathrm{x}\right)>{\log }_{10}(\mathrm{x}-4)\left(4\right)$$$$\mathrm{From}\left(2\right)\left(3\right)\left(4\right)\mathrm{we}\mathrm{get}$$$${\log }_2(10+2\mathrm{x})-2>{\log }_{10}(\mathrm{x}-4).\mathrm{Hence}$$$$\mathrm{LHS}\left(1\right)>0\mathrm{which}\mathrm{means}\mathrm{the}\mathrm{equation}$$$$\mathrm{has}\mathrm{no}\mathrm{solutions}$$$$\mathbf{Thus},\mathbf{there}{\mathbf{don}}^{\prime }\mathbf{t}\mathbf{exist}\mathbf{x}\mathbf{satisfying}$$$$\mathbf{the}\mathbf{equality}$$$${\mathbf{log}}_2(10+2\mathbf{x})={\mathbf{log}}_{10}(\mathbf{x}-4)+2$$

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