Question 997511: I am stuck on this one. Any help will be appreciated! e^x-4e^(-x)=3 x=ln4
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! start with e^x - 4e^(-x) = 3
since e^(-x) is the equivalent to 1/e^x, the equation becomes:
e^x - 4/(e^x) = 3
multiply both sides of this equation by e^x and you get:
e^x * e^x - 4 = 3 * e^x
e^x * e^x is the same as e^(x+x) is the same as e^(2x) which is the same as (e^x)^2.
your equation becomes (e^x)^2 - 4 = 3 * e^x
subtract 3 * e^x from both sides of this equation to get (e^x)^2 - 3e^x - 4 = 0
let y = e^x and the equation becomes y^2 - 3y - 4 = 0
this is a quadratic equation that can be factored to get (y - 4) * (y + 1) = 0
solve for y to get y = 4 or y = -1.
since y = e^x, replace y to get:
e^x = 4 or e^x = -1
e^x = 4 if and only if ln(4) = x
this matches the solution you showed, so this is the correct answer.
e^x = -1 if and only if ln(-1) = x
since you cannot take the log of a negative number, this solution is invalid.
your only solution is x = ln(4).
you could also have stopped at e^x = -1 since y = e^x is only valid for positive values of y.
|
|
|
