Hello Caitlin,

You can turn this equation into an equation that is quadratic in form by multiplying both sides by e^{x}. That is,

e^{x} + 45e^{-x} - 14 = 0

e^{x}(e^{x} + 45e^{-x} - 14) = e^{x}⋅0

Distribute and add exponents in the above equation to get

e^{2x} + 45e^{0} - 14e^{x} = 0

e^{2x} + 45 - 14e^{x} = 0 (using the fact that e^{0} = 1.)

e^{2x} - 14e^{x} + 45 = 0.

Now substitute u = e^{x}. Then, since u^{2} = (e^{x})^{2} = e^{2x}, the above equation can be rewritten as

u^{2} - 14u + 45 = 0.

(Since the original equation can be turned into a quadratic equation by using an appropriate substitution, we say that the original equation is **quadratic in form**.) Now it is easy to solve the last equation by factoring.

(u - 9)(u - 5) = 0

u - 9 = 0 or u - 5 = 0

u = 9 or u = 5.

To complete the solution, we substitute e^{x} back in for u, and then take the natural log of each side.

e^{x} = 9 or e^{x} = 5.

Ln(e^{x}) = Ln(9) or Ln(e^{x}) = Ln(5)

x = Ln(9) or x = Ln(5)

**x ≅ 2.1972 or x ≅ 1.6094** (rounded to 4 decimal places.)

Hope that helps! Let me know if you need any further help.

William