**Real roots**

This is a quadratic in #e^x#. It will be easier to see what we're doing if we introduce a substitution. I will let #t=e^x#:

#(e^x)^2-6e^x-16=0#

#t^2-6t-16=0#

We solve the quadratic by grouping:

#t^2-8t+2t-16=0#

#t(t-8)+2(t-8)=0#

#(t+2)(t-8)=0#

By the zero factor principle, this has solutions #t=-2# and #t=8#. We can undo the substitution to get these in terms of #x#:

#e^x=8#

#ln(e^x)=ln(8)#

#x=ln(8)#

**Complex roots**

For the other root, #t=-2#, we can try doing the same procedure:

#e^x=-2#

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

#x=ln(-2)#

But we run into a problem. The natural log of negative numbers is not defined with real numbers, which means that this root is complex.

We can find this root by using Euler's identity, #e^(ipi)=-1#. This lets us rewrite the logarithm as such:

#ln(-2)=ln(2*-1)=ln(2e^(ipi))=ln(2)+ln(e^(ipi))=ln(2)+ipi#

Note that #ipi# is not the only imaginary exponent that gives #-1#. Euler's formula actually gives an infinite number of ways to express #-1# in this way due to the periodicity of the sine and cosine functions. We can capture this by adding #2pik# (remember, the periodicity of sine and cosine is #2pi#) in the exponent, which after the same calculations as before gives the solutions:

#x=ln(2)+(2k+1)ipi#

Using complex numbers, we can also find even more solutions!

Because Euler's identity gives that #e^(i2pik)=1#, we can do the same thing we did with #ln(-2)# with #ln(8)#:

#ln(8)=ln(8e^(i2pik))=ln(8)+ln(e^(i2pik))=ln(8)+i2pik#