How to solve this? e^(2x+1) + 2e^(x+2) - 3 = 0
Solve this by turning it into a quadratic equation.
Rearrange the equation as follows,
e^(2x+1) + 2e^(x+2) - 3 = 0
e^(2x+4) / e^3 + 2e^(x+2) - 3 = 0
e^2(x+2) / e^3 + 2e^(x+2) - 3 = 0
e^2(x+2) + 2.e^3.e^(x+2) – 3.e^3 = 0
u^2 + 2.e^3.u – 3.e^3 = 0, where u = e^(x+2)
Now use the quadratic formula to solve the quadratic.
u = (-2.e^3 +/- sqrt(4e^6 – 4.(1).(-3.e^3))) / (2*1)
u = (-e^3 +/- sqrt(e^6 + 3.e^3))
u = e^3(-1 +/- sqrt(1 + 3/e^3))
Evaluating this,
u = -41.61889, 1.44782
ignoring the negative result, since x = ln(u) – 2, from the earlier substitution, and we cannot take the log of a negative number. Then
x = ln(1.44782) – 2
x = 0.370058 – 2
x = -1.62994