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Question Number 194434 by SANOGO last updated on 06/Jul/23
calcul  e^(2ln(1+u) ) −e^(−2ln(1+u))  =?
$${calcul} \\ $$$${e}^{\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)\:} −{e}^{−\mathrm{2}{ln}\left(\mathrm{1}+{u}\right)} \:=? \\ $$
Answered by aba last updated on 06/Jul/23
e^(2ln(1+u)) −e^(−2ln(1+u)) =e^(ln(1+u)^2 ) −e^(ln((1/(1+u)))^2 )                                           =(1+u)^2 −(1/((1+u)^2 ))
$$\mathrm{e}^{\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)} −\mathrm{e}^{−\mathrm{2ln}\left(\mathrm{1}+\mathrm{u}\right)} =\mathrm{e}^{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} } −\mathrm{e}^{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{u}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{2}} } \\ $$

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