Find, as natural logarithms, the roots of the equation 2e^x + 3e^-x = 7. Can anyone help?

Answer 1

let y =e^x
2y + 3/y = 7
2y^2 - 7y + 3 = 0
factor that.
Choose only positive y since y = e^x.
Now x = ln y.

d:

Answer 2

2e^x + 3e^(-x) = 7

Let y = e^x
Notice that 1/y = e^(-x)

So the equation becomes:
2y + 3/y = 7

2y^2 + 3 = 7y

2y^2 - 7y + 3 = 0

Quadratic formula: y = (7 +/- sqrt(49-24))/4
= ( 7 +/- sqrt(25))/4
= (7 +/- 5)/4
= 3, 1/2.

so the answers are
e^x = y = 3 or 1/2,

which means
x = ln(3) or ln(1/2) = -ln(2)

Answer 3

substitute e^x = z (some arbitrary variable z)
then ur equation
2e^x + 3e^-x = 7 can be rewritten as
2z + 3/z = 7
or simplifying,
2z^2 - 7z + 3 = 0
solve this quadratic in z
(2z - 1)(z - 3) = 0

its roots are z = 1/2 & z = 3

so the solution to ur original eqn. is
e^x = 1/2 & e^x = 3,

or in natural logarithms,
x = ln(1/2) & x = ln(3)

Answer 4

Remember, e^-x = 1/(e^x)

Multiply both sides by e^x:
2e^2x + 3 = 7e^x

2e^2x - 7e^x + 3 = 0

Take t=e^x

2t^2 - 7t + 3 = 0

t1,2 = [7 +- sqrt(49 - 2*3*4)]/4 = [7 +- sqrt(25)]/4 =
= (7 +- 5)/4

t1 = 3: x1 = ln 3

t2 = 1/2: x2=ln(1/2) = ln1 - ln2 = -ln 2

Answer 5

2e^x + 3e^-x = 7

2e^x + 3/e^x = 7

(2e^x * e^x)/e^x + 3/e^x = 7

2(e^x)^2 / e^x + 3/e^x = 7

(2e^(2x) + 3) / e^x = 7

2e^(2x) + 3 = 7e^x


let y be e^x

2y^2 + 3 = 7y

2y^2 - 7y + 3 = 0

2y^2 - 6y - 1y + 3 = 0

(2y^2 - 6y) + (-1y + 3) = 0

2y(y - 3) + -1(y - 3) = 0

2y(y - 3) - 1(y - 3) = 0

(2y - 1) (y - 3) = 0

y = 1/2 or 3

we know that y = e^x

1/2 = e^x
x = ln(1/2)

3 = e^x
x = ln(3)

so x = ln(1/2) or ln(3)

Answer 6

2e^x + 3e^(-x) = 7
2e^(2x) + 3 = 7 e^x
2e^(2x) - 7e^x + 3 = 0
(2e^x - 1).(e^x - 3) = 0
e^x = 1 /2 , e^x = 3
x = ln(1/2) , x = ln 3

Answer 7

call e^x=z so
2z+3/z=7
2z^2-7z+3=0
z=(7+-sqrt(49-24)))/4 so z= 3 and z= 1/2
e^x=3 so x = ln3 and e^x=1/2 and x= ln(1/2)=-ln2

Answer 8

2e^x + 3e^-x = 7 /*e^x
2e^2x+3=7e^x
2e^2x-7e^x+3=0
e^x=t
2t^2-7t+3=0
t1=1/2 t2=3
e^x=1/2=> x1=ln1/2 <=> x1=-ln2
e^x=3=> x2=ln3