What is e^2x-e^x-12=0? We are solving for x and logs are required in the solution.

Answer 1

e^2x - e^x -12 = 0

let e^x=y,

then e^2x=e^x(e^x) = y^2

y^2 -y -12 = 0

factor

(y-4)(y+3) = 0

set each factor =0

y-4=0 or y+3=0

y=4 or y =-3

e^x =4

x = ln4

x = about 1.3863

e^x =-3

x = ln(-3) but there are no real natural logs of negative numbers

that's where you'd stop in a normal math course problem

but maybe

you could use Euler's identity. maybe not

eipi = -1

take the natural log of both sides

ipi = ln(-1)

ln(-1) = ipi which is an imaginary number

so is there an imaginary soluton to ln(-3)?

that seems to require the right side being =-3, but consines are never less than -1

but ln(-3) = ln(-1)+ln(3) = ipi +ln3, another complex number

lna +lnb = lnab

so x = ln4 or ln3+ipi

doubtful any instructor or textbook would expect this as the answer,

they want just the real solution

check the solution, x=ln4

e^2ln4 - e^ln4 -12

= e^ln16 - 4-12

=16-16

= 0

x=ip+ln3

e^(2ip+2ln3) - e^(ip+ln3) -12

= (e^2ip)(e^ln9) - (e^ip)(e^ln3) -12

= 9e^2ip -3e^ip -12

=3(e^2ip-e^ip)-12

which would have to be

=3(4)-12

=0

but e^2ip-e^ip = e(2ip/ip) = e^2 = about 7.39

e^ln4 = 4. ln4 =about 1.39

a "tiny" arithmetic mistake somewhere

that similar .39 two digits in both look promising though