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

lg^2 (10x) + lg(10x) = 6 - lg(x) x = ?

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

Answered by Olaf_Thorendsen last updated on 26/Jul/21

lg^2 (10x)+lg(10x) = 6−lg(x) lg^2 (10x)+lg(10x)+lg(x)−6 = 0 lg^2 (10x)+lg(10x)+lg(10x)−6−lg(10) = 0 lg^2 (10x)+2lg(10x)−7 = 0 lg(10x) = ((−2±(√(4−4(−7))))/2) lg(10x) = −1±2(√2) 10x = 10^(−1±2(√2)) x = 10^(−2±2(√2))

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

Commented by mathdanisur last updated on 27/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

Answered by liberty last updated on 27/Jul/21

⇒(1+log x)^2 +(1+log x)=6−log x ⇒(1+t)^2 +2t−5=0 ⇒t^2 +4t−4=0 ⇒(t+2)^2 = 8 ⇒t+2 = ±2(√2) ⇒t=−2±2(√2) ⇒log x=−2±2(√2) ⇒x=10^(−2±2(√2))

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

Commented by mathdanisur last updated on 27/Jul/21

thank you Ser

$${thank}\:{you}\:{Ser} \\ $$

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