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Question Number 148300 by mathdanisur last updated on 26/Jul/21

$${lg}^2\left(10x\right)+lg\left(10x\right)=6-lg\left(x\right)$$$$x=?$$

Answered by Olaf_Thorendsen last updated on 26/Jul/21

$${\lg }^2\left(10x\right)+\lg \left(10x\right)=6-\lg \left(x\right)$$$${\lg }^2\left(10x\right)+\lg \left(10x\right)+\lg \left(x\right)-6=0$$$${\lg }^2\left(10x\right)+\lg \left(10x\right)+\lg \left(10x\right)-6-\lg \left(10\right)=0$$$${\lg }^2\left(10x\right)+2\lg \left(10x\right)-7=0$$$$\lg \left(10x\right)=\frac{-2\pm \sqrt{4-4(-7)}}{2}$$$$\lg \left(10x\right)=-1\pm 2\sqrt{2}$$$$10x={10}^{-1\pm 2\sqrt{2}}$$$$x={10}^{-2\pm 2\sqrt{2}}$$

Commented by mathdanisur last updated on 27/Jul/21

$$thankyouSer$$

Answered by liberty last updated on 27/Jul/21

$$\Rightarrow {(1+\log \mathrm{x})}^2+(1+\log \mathrm{x})=6-\log \mathrm{x}$$$$\Rightarrow {(1+\mathrm{t})}^2+2\mathrm{t}-5=0$$$$\Rightarrow {\mathrm{t}}^2+4\mathrm{t}-4=0$$$$\Rightarrow {(\mathrm{t}+2)}^2=8$$$$\Rightarrow \mathrm{t}+2=\pm 2\sqrt{2}$$$$\Rightarrow \mathrm{t}=-2\pm 2\sqrt{2}$$$$\Rightarrow \log \mathrm{x}=-2\pm 2\sqrt{2}$$$$\Rightarrow \mathrm{x}={10}^{-2\pm 2\sqrt{2}}$$

Commented by mathdanisur last updated on 27/Jul/21

$$thankyouSer$$

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