Solve for x: e^(2x) - 7e^(x) + 12 = 0?

Question

Give exact answers

Answer 1

Using ln substitution method and (requires calculator):
Let y = e to the power of x
Therefore, y square - 7 y + 12 = 0
(y - 3) (y - 4) = 0
y = 3 or y = 4
Thus, e to the power of x = 3 or 4
x = ln 3 or x = ln 4
x = 1.10 or x = 1.39 (giving your answers to 3 sig. fig.)

Answer 2

Factor the left-hand side:
(e^x -3)(e^x -4) = 0.
Now set each factor to zero and solve:
e^x = 3, so x = ln 3
e^x = 4, so x = ln 4

Answer 3

Note that e^(2x) = (e^x)^2
from the general power rules, i.e (a^x)^n = a^nx

so this is a quadratic equation
let y= e^x then
y^2 - 7y + 12 = 0
y= [7+/- V(49- 4.1.12)]/2 = 4 or 3
then e^x = 4 or 3
x = ln 4 or ln 3

et voila!

Answer 4

e^(2x) - 7e^x + 12 = 0
(e^x)^2 - 7(e^x) + 12 = 0
(e^x - 4)(e^x - 3) = 0

e^x - 4 = 0
e^x = 4
x = ln(4)

e^x - 3 = 0
e^x = 3
x = ln(3)

ANS : x = ln(3) or ln(4)

Answer 5

let t = e^x
then eq is
t^2-7t+12=0
it is quardratic
(t-3)(t-4)
t=3
t=4
e^x=3
e^x=4
x=ln3
x=ln 4

Answer 6

Let u=e^x,

u^2-7u+12=0
(u-3)(u-4)=0
u-3=0 or u-4=0
u=3 or u=4
e^x=3 or e^x=4
lne^x=ln3 or lne^x=ln4
x=ln3 or x=ln4
x=1.10 (3s.f.) or x=1.39(3s.f.)

Answer 7

never went pass consumer math lol....