using suitable substuitutions, solve for x. 2e^x = 7 square root e^x - 3?

Question

answer -1.39 and 2.20

Answer 1

2e^x=7 sqrt(e^x-3)
1)e^x-3>=0 so e^x>=3
Put e^x= z
2z=7 sqrt(z-3) squaring both sides
4z^2=49z -147
4z^-49z+147 = 0
z=((49+-sqrt(49)/8 =
z=7 and z= 42/8=21/4
e^x=7 so x = ln7
e^x=21/4 so x= ln(21/4)
If you meant
2e^x=7sqrt(e^x)-3
2z+3=7sqrt(z)
4z^2+12z +9 = 49z
4z^2-37z +9 =0
z=((37+-35)/8

z=9 and z= 1/4
x= ln9 =2.20 and x= ln(1/4)=-1.39

Answer 2

let t = sqrt[e^x], condition: t > 0.

2 e^x = 7 * sqrt[e^x] - 3
<=> 2t^2 = 7 t - 3
<=> (quadratic eq.) 2t^2 - 7t + 3= 0

=> t = 1/2 or t = 3
=> x = ln(t^2) = 2ln(t)
x1 = -1.39 or x = 2.20

Answer 3

Substitute y = e^x
2y = 7sqrt(y - 3)... square everything
4y² = 49(y - 3)

Solve the quadratic equation, then x = ln y.