Respuesta :
Answer:
1) The equation rewritten as a quadratic equation in standard form using u-substitution is:
2u² + 7u - 15 = 0
2) The exact solution is:
x = 0.41
Explanations:The qiven equation is:
[tex]2e^{2x}+7e^x\text{ = 15}[/tex]This can be rewritten as:
[tex]2(e^x)^2+7e^x\text{ - 15 = 0}[/tex][tex]\text{Let u = e}^x[/tex]The equation rewritten as a quadratic equation in standard form becomes:
[tex]2u^2\text{ + 7u - 15 = 0}[/tex]Factorize the quadratic equation above:
[tex]\begin{gathered} 2u^2\text{ - 3u + 10u - 15 = 0} \\ u\text{ (2u - 3) + 5(2u - 3) = 0} \\ (u\text{ + 5) (2u - 3) = 0} \\ u\text{ + 5 = 0} \\ u\text{ = -5} \\ 2u\text{ - 3 = 0} \\ 2u\text{ = 3} \\ u\text{ = }\frac{3}{2} \end{gathered}[/tex]u = -5 or u = 3/2
[tex]\begin{gathered} \text{But e}^x\text{ = u} \\ e^{x\text{ }}=-5^{} \\ x\text{ = ln(-5) \lbrack{}Invalid solution\rbrack} \\ e^x\text{ = 3/2} \\ e^{x\text{ }}\text{ = 1.5} \\ x\text{ = ln(1.5)} \\ x\text{ = }0.41 \end{gathered}[/tex]