What is the quadratic formula of #e^(2x) - 2e^x = 1#?

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Answer 1 · 2015-10-31T21:33:40

Recognise this as quadratic in #e^x# and hence solve using the quadratic formula to find:

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

This is an equation that is quadratic in #e^x#, rewritable as:

#(e^x)^2-2(e^x)-1 = 0#

If we substitute #t = e^x#, we get:

#t^2-2t-1 = 0#

which is in the form #at^2+bt+c = 0#, with #a=1#, #b=-2# and #c=-1#.

This has roots given by the quadratic formula:

#t = (-b+-sqrt(b^2-4ac))/(2a) = (2+-sqrt(4+4))/2 = 1+-sqrt(2)#

Now #1-sqrt(2) < 0# is not a possible value of #e^x# for Real values of #x#.

So #e^x = 1 + sqrt(2)# and #x = ln(1+sqrt(2))#