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Question Number 207717 by hardmath last updated on 24/May/24

$${\lg }^2\left(10\mathrm{x}\right)-\lg 10\mathrm{x}+1=6-\lg \mathrm{x}$$$$\mathrm{find}:\mathbf{x}=?$$

Answered by TonyCWX08 last updated on 24/May/24

$${(1+lg\left(x\right))}^2-(1+lg\left(x\right))+lg\left(x\right)+1-6=0$$$$1+2lg\left(x\right)+{lg}^2\left(x\right)-1-lg\left(x\right)+lg\left(x\right)-5=0$$$${lg}^2\left(x\right)+2lg\left(x\right)-5=0$$$$lety=lg\left(x\right)$$$$y^2+2y-5=0$$$$y=-1\pm \sqrt{6}$$$$lg\left(x\right)=-1\pm \sqrt{6}$$$$x={10}^{-1\pm \sqrt{6}}$$$$$$$$x_1={10}^{-1+\sqrt{6}}$$$$x_2={10}^{-1-\sqrt{6}}$$

Commented by hardmath last updated on 24/May/24

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}$$

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