Question #1f1cc

2 Answers
Jim S
Jan 30, 2018

#x=-ln(e-1)/2#

Explanation:

#e-e^(-2x)=1#

#e^(-2x)=e-1#

#lne^(-2x)=ln(e-1)#

#-2x*lne=ln(e-1)#

#-2x*1=ln(e-1)#

#-2x=ln(e-1)#

#x=-ln(e-1)/2#

Barry H.
Jan 30, 2018

#x=ln[1/[sqrt[e-1]]]#

Explanation:

#e- e^[-2x]# can be written as #e-[e^-x]^2# so the expression becomes #-e-[e^-x]^2=1#, tidying up both sides gives #[e^-x]^2=[e-1]#, and then taking the sqare root of both sides,

#e^-x=sqrt [e-1]#, and therefore #e^x=1/sqrt[e-1]#.

Taking logs of both sides, # lne^x=ln[1/sqrt[e-1]]# #but ln [e^x]=x# and so #x=ln[1/sqrt[e-1]]#. Hope this helps.

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