exponential question

Question

e^2x+1 + 9e^x - 11 = 0

Answer 1

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·e¹ = e·e^2x.

Then your equation can be written

e·e^2x + 9e^x - 11 = 0

Now substitute u = e^x. Then, since u² = (e^x)² = e^2x, the above equation can be rewritten as

eu² + 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 ±√(b² -4ac)]/2a

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

u = [-9 ±√(9² - 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

Answer 2

e^2x+1 + 9e^x - 11 = 0

e(e^x)² + 9(e^x) - 11 = 0

By the Quadratic Formula, e^x = [-9 ± √(81 + 44e)] / (2e) = [-9 ± √200.604] / (2e)

e^x = 0.9498 or -4.26

Since e^x is always positive, e^x = 0.9498

So, x = ln(0.9498) = -0.0515

Answer 3

The best I can do is x=1*10^-100=11.71828186.....

-11=.718