Hello Mayowa,

An equation like the one you present can often be transformed into an equation that is **quadratic in form** using an appropriate substitution. Before proceeding with the substitution, note that

**e**^{2x+1}** = e**^{2x}**·e**^{1}** = e·e**^{2x}.

Then your equation can be written

e·e^{2x} + 9e^{x} - 11 = 0

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

eu^{2} + 9u - 11 = 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**.) We will use the quadratic equation to solve for u.

u = [-b ±√(b^{2} -4ac)]/2a

Here, a = e, b = 9, and c = -11,

u = [-9 ±√(9^{2} - 4e(-11))]/(2e)

u = [-9 ±√(81+ 44e)]/(2e)

To complete the solution, we substitute e^{x} back in for u and solve for x.

e^{x} = [-9 +√(81 + 44e)]/(2e) or e^{x} = [-9 -√(81 + 44e)]/(2e)

Recall that e^{x} is positive for all x. You can check that [-9 -√(81 + 44e)]/(2e) is negative, so we discard the second possible solution above, and the only possible solution will be obtained from

e^{x} = [-9 +√(81 + 44e)]/(2e)

Now take the natural logarithm of each side.

Ln(e^{x}) = Ln{[-9 +√(81 + 44e)]/(2e)}

**x = Ln{[-9 +√(81 + 44e)]/(2e)}** (Exact form)

**x≅-0.05153** (Rounded to the 5th decimal place.)

Hope that helps! Let me know if you need any additional explanation.

William