How do you solve #e^(2x)+8e^x+12 = 0# ?

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Answer 1 · 2015-06-21T21:43:38

This has no real solutions, but if you allow complex values it has solutions:

#x = log_e 2 + i(2n+1)pi# for all #n in ZZ#

and

#x = log_e 6 + i(2n+1)pi# for all #n in ZZ#

#0 = e^(2x)+8e^x+12 = (e^x + 6)(e^x + 2)#

So we are looking for #x# such that #e^x = -2# or #e^x = -6#.

There are no real values of #x# that satisfy this, but we can use #e^(ipi) = -1# and the properties of exponents to find solutions:

#x = log_e 2 + i(2n+1)pi# for all #n in ZZ#

and

#x = log_e 6 + i(2n+1)pi# for all #n in ZZ#

For example,

#e^(log_e 2 + i(2n+1)pi)#

#= e^(log_e 2)*(e^(i pi))^(2n+1)#

#= 2*(-1)^(2n+1)#

#= -2#