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Question Number 4478 by love math last updated on 30/Jan/16

log_(10) (x^2 +1)=(2/(log_(10) (x^2 +1)))−2

$${log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)=\frac{\mathrm{2}}{{log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)}−\mathrm{2} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by Rasheed Soomro last updated on 30/Jan/16

[log_(10) (x^2 +1)]^2 =2−2(log_(10) (x^2 +1)) [log_(10) (x^2 +1)]^2 +2(log_(10) (x^2 +1))−2=0 let log_(10) (x^2 +1)=y y^2 +2y−2=0 y=((−2±(√(2^2 −4(1)(−2))))/(2(1))) y=((−2±(√(12)))/2)=−1±(√3) log_(10) (x^2 +1)=−1±(√3) x^2 +1=10^(−1±(√3)) x=±(√(10^(−1±(√3)) −1)) {(√(10^(−1+(√3)) −1)) , (√(10^(−1−(√3)) −1)) ,−(√(10^(−1+(√3)) −1)) ,(√(10^(−1−(√3)) −1))}

$$\left[{log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]^{\mathrm{2}} =\mathrm{2}−\mathrm{2}\left({log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\right) \\ $$$$\left[{log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]^{\mathrm{2}} +\mathrm{2}\left({log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\right)−\mathrm{2}=\mathrm{0} \\ $$$${let}\:{log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)={y} \\ $$$${y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{2}=\mathrm{0} \\ $$$${y}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(−\mathrm{2}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$${y}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{12}}}{\mathrm{2}}=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$$\:\:\:{log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{10}^{−\mathrm{1}\pm\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:{x}=\pm\sqrt{\mathrm{10}^{−\mathrm{1}\pm\sqrt{\mathrm{3}}} −\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\left\{\sqrt{\mathrm{10}^{−\mathrm{1}+\sqrt{\mathrm{3}}} −\mathrm{1}}\:,\:\sqrt{\mathrm{10}^{−\mathrm{1}−\sqrt{\mathrm{3}}} −\mathrm{1}}\:,−\sqrt{\mathrm{10}^{−\mathrm{1}+\sqrt{\mathrm{3}}} −\mathrm{1}}\:,\sqrt{\mathrm{10}^{−\mathrm{1}−\sqrt{\mathrm{3}}} −\mathrm{1}}\right\} \\ $$

Commented by RasheedSindhi last updated on 30/Jan/16

Continue from answer y=−1±(√3) x^2 +1=−1±(√3) x^2 =−2±(√3) x=±(√(−2+(√3))) ,±(√(−2−(√3))) All roots are complex , because numbers under (√( )) are negative.

$${Continue}\:{from}\:{answer} \\ $$$${y}=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} =−\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$${x}=\pm\sqrt{−\mathrm{2}+\sqrt{\mathrm{3}}}\:,\pm\sqrt{−\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${All}\:{roots}\:{are}\:{complex}\:,\:{because} \\ $$$${numbers}\:{under}\:\sqrt{\:\:\:}\:\:{are}\:{negative}. \\ $$$$ \\ $$

Commented by FilupSmith last updated on 31/Jan/16

You wrote above: x^2 +1=−1±(√3) Is this correct? It was said that y=log_(10) (x^2 +1) ∴log_(10) (x^2 +1)=−1±(√3) x^2 =10^(−1±(√3)) −1 ∴x=±(√(10^(−(1−(√3))) −1)), ±(√(10^(−(1+(√3))) −1)) Have I misread your working?

$$\mathrm{You}\:\mathrm{wrote}\:\mathrm{above}: \\ $$$${x}^{\mathrm{2}} +\mathrm{1}=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}? \\ $$$$ \\ $$$$\mathrm{It}\:\mathrm{was}\:\mathrm{said}\:\mathrm{that}\:{y}=\mathrm{log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\therefore\mathrm{log}_{\mathrm{10}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$${x}^{\mathrm{2}} =\mathrm{10}^{−\mathrm{1}\pm\sqrt{\mathrm{3}}} −\mathrm{1} \\ $$$$ \\ $$$$\therefore{x}=\pm\sqrt{\mathrm{10}^{−\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)} −\mathrm{1}},\:\pm\sqrt{\mathrm{10}^{−\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)} −\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{Have}\:\mathrm{I}\:\mathrm{misread}\:\mathrm{your}\:\mathrm{working}? \\ $$

Commented by RasheedSindhi last updated on 30/Jan/16

Thanks! It′s a mistake. You are very right! I am going to correct my answer.

$${Thanks}!\:{It}'{s}\:{a}\:{mistake}. \\ $$$${You}\:{are}\:{very}\:{right}!\:{I}\:{am}\:{going}\:{to} \\ $$$${correct}\:{my}\:{answer}. \\ $$

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