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Question Number 178575 by Spillover last updated on 18/Oct/22
Solve for x  e^(sinh^(−1) x) =1+e^(cosh^(−1) x)
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x} \\ $$$$\mathrm{e}^{\mathrm{sinh}\:^{−\mathrm{1}} \mathrm{x}} =\mathrm{1}+\mathrm{e}^{\mathrm{cosh}\:^{−\mathrm{1}} \mathrm{x}} \\ $$
Answered by mr W last updated on 19/Oct/22
x+(√(x^2 +1))=1+x+(√(x^2 −1))  (√(x^2 +1))=1+(√(x^2 −1))  x^2 +1=x^2 +2(√(x^2 −1))  1=2(√(x^2 −1))  x^2 −1=(1/4)  ⇒x=±((√5)/2)
$${x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${x}^{\mathrm{2}} +\mathrm{1}={x}^{\mathrm{2}} +\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mathrm{1}=\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${x}^{\mathrm{2}} −\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow{x}=\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Commented by Spillover last updated on 19/Oct/22
thanks
$$\mathrm{thanks} \\ $$

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