Solve for x e^(2x) - 4(e^x) - 5 = 0
N nicole New member Joined Feb 3, 2005 Messages 2 Oct 28, 2005 #1 Solve for x e^(2x) - 4(e^x) - 5 = 0
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Oct 28, 2005 #2 Hello, nicole! Solve for x: e<sup>2x</sup> - 4e<sup>x</sup> - 5 = 0 Click to expand... Note that we have: .(e<sup>x</sup>)<sup>2</sup> - 4(e<sup>x</sup>) - 5 .= .0 . . . a quadratic. Let u = e<sup>x</sup> Then we have: .u<sup>2</sup> - 4u - 5 .= .0 . . which factors: .(u - 5)(u + 1) .= .0 . . and has roots: .u = 5, -1. Since u = e<sup>x</sup>, we have: .e<sup>x</sup> = 5 . ---> . x = ln(5). . . . . . . . . . . . . . . .and: .e<sup>x</sup> = -1 . . . which has no real root Therefore, the only answer is: .x = ln(5)
Hello, nicole! Solve for x: e<sup>2x</sup> - 4e<sup>x</sup> - 5 = 0 Click to expand... Note that we have: .(e<sup>x</sup>)<sup>2</sup> - 4(e<sup>x</sup>) - 5 .= .0 . . . a quadratic. Let u = e<sup>x</sup> Then we have: .u<sup>2</sup> - 4u - 5 .= .0 . . which factors: .(u - 5)(u + 1) .= .0 . . and has roots: .u = 5, -1. Since u = e<sup>x</sup>, we have: .e<sup>x</sup> = 5 . ---> . x = ln(5). . . . . . . . . . . . . . . .and: .e<sup>x</sup> = -1 . . . which has no real root Therefore, the only answer is: .x = ln(5)
