How do you solve for X?

Question

#e^(2-2ln(x²+1))=1#

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Answer 1 · 2018-05-19T01:29:18

#color(blue)(x = +- sqrt(e-1)#

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

#ln e^(2-2 ln(x^2+1)) = ln 1#, taking ln on both sides

#(2-2ln (x^2+1)) = 0# as ln 1 = 0#

#2 ln (x^2+1) = 2#

#ln (x^2 + 1) = 1#

#x^2 + 1 = e^1 = e#

#x^2 = e-1#

#color(blue)(x = +- sqrt(e-1)#