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Question Number 35743 by mondodotto@gmail.com last updated on 22/May/18

find x log(log2+log_2 (x+1))=0

$$\boldsymbol{\mathrm{find}}\:\boldsymbol{{x}} \\ $$$$\boldsymbol{\mathrm{log}}\left(\boldsymbol{\mathrm{log}}\mathrm{2}+\boldsymbol{\mathrm{log}}_{\mathrm{2}} \left(\boldsymbol{{x}}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$

Commented by prof Abdo imad last updated on 25/May/18

if log mean ln (e)⇔ ln(2) +ln_2 (x+1) =1 ⇔ ln(2) + ((ln(x+1))/(ln(2))) =1 with x>−1 ⇔{ln(2)}^2 +ln(x+1) = ln(2) ⇔ln(x+1)=ln(2) −{ln(2)}^2 ⇔ x+1 =e^(ln(2) −(ln(2))^2 ) =2 e^(−{ln(2)}^2 ) ⇔ x = 2 e^(−{ln(2)}^2 ) −1 .

$${if}\:{log}\:{mean}\:{ln}\:\:\left({e}\right)\Leftrightarrow\:{ln}\left(\mathrm{2}\right)\:+{ln}_{\mathrm{2}} \left({x}+\mathrm{1}\right)\:=\mathrm{1} \\ $$$$\Leftrightarrow\:{ln}\left(\mathrm{2}\right)\:+\:\frac{{ln}\left({x}+\mathrm{1}\right)}{{ln}\left(\mathrm{2}\right)}\:=\mathrm{1}\:\:\:{with}\:{x}>−\mathrm{1} \\ $$$$\Leftrightarrow\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} \:+{ln}\left({x}+\mathrm{1}\right)\:=\:{ln}\left(\mathrm{2}\right) \\ $$$$\Leftrightarrow{ln}\left({x}+\mathrm{1}\right)={ln}\left(\mathrm{2}\right)\:−\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} \:\Leftrightarrow \\ $$$${x}+\mathrm{1}\:={e}^{{ln}\left(\mathrm{2}\right)\:−\left({ln}\left(\mathrm{2}\right)\right)^{\mathrm{2}} } \:=\mathrm{2}\:{e}^{−\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} } \:\Leftrightarrow \\ $$$${x}\:=\:\mathrm{2}\:{e}^{−\left\{{ln}\left(\mathrm{2}\right)\right\}^{\mathrm{2}} } \:−\mathrm{1}\:\:. \\ $$

Answered by MJS last updated on 23/May/18

log=ln I hope... ln(ln(2)+log_2 (x+1))=0 ln(2)+log_2 (x+1)=1 log_2 (x+1)=1−ln(2) (x+1)=2^(1−ln(2)) x=(2/2^(ln(2)) )−1≈.237

$$\mathrm{log}=\mathrm{ln}\:\mathrm{I}\:\mathrm{hope}... \\ $$$$\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)\right)=\mathrm{0} \\ $$$$\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{log}_{\mathrm{2}} \left({x}+\mathrm{1}\right)=\mathrm{1}−\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\left({x}+\mathrm{1}\right)=\mathrm{2}^{\mathrm{1}−\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$${x}=\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{ln}\left(\mathrm{2}\right)} }−\mathrm{1}\approx.\mathrm{237} \\ $$

Commented by mondodotto@gmail.com last updated on 23/May/18

thanx

$$\mathrm{thanx} \\ $$

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