e^x - e^(-x) = 1 — step-by-step solution

1) Let's solve in

ℝ

e^{x} - e^{- x} = 1

ⓘ The factor is moved to the numerator or the denominator by changing the sign of its exponent, for a nonzero base:

e^{- x}

\frac{1}{e^{x}}

e^{x} - \frac{1}{e^{x}} = 1

We multiply both sides by

e^{x}

, which is strictly positive

e^{x} \cdot e^{x} - \frac{1}{e^{x}} \cdot e^{x} = e^{x}

ⓘ Multiplying powers with the same base adds the exponents:

e^{x} \cdot e^{x}

e^{x + x}

■Multiplying a fraction by a term multiplies its numerator:

\frac{1}{e^{x}} \cdot e^{x}

\frac{1 \cdot e^{x}}{e^{x}}

\frac{e^{x}}{e^{x}}

■The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

- \frac{e^{x}}{e^{x}}

-1

e^{x + x} - 1 = e^{x}

ⓘ Like terms are grouped by adding their coefficients:

x + x

2 x

e^{2 x} - 1 = e^{x}

Let's define:

X = e^{x}

You can rephrase the expression into:

X^{2} - X - 1 = 0

Given the quadratic form

a X^{2} + b X + c

, let's compute the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = -1

and

c = -1

Δ = {(-1)}^{2} - 4 \cdot 1 \cdot (-1)

ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product:

{(-1)}^{2}

(-1) \cdot (-1)

1

Δ = 1 - 4 \cdot (-1)

ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive:

- 4 \cdot (-1)

4

Δ = 1 + 4

ⓘ Start from

1

and move forward

4

on the number line:

1 + 4

5

Δ = 5

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

X_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- (-1) - \sqrt{5}}{2 \cdot 1}

X_{1} = \frac{1 - \sqrt{5}}{2}

The quadratic formula gives

X_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- (-1) + \sqrt{5}}{2 \cdot 1}

X_{2} = \frac{1 + \sqrt{5}}{2}

Solutions obtained:

X = {\frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2}}

2) Solving for x:

X = e^{x}

using

X = \frac{1 - \sqrt{5}}{2}

e^{x} = \frac{1 - \sqrt{5}}{2}

In ℝ, the exponential function is strictly positive:

e^{x} > 0

, whereas

\frac{1 - \sqrt{5}}{2}

\approx -0.6180

is negative, so

e^{x} = \frac{1 - \sqrt{5}}{2}

has no solution

Solution:

x \in \emptyset

3) Let's solve for x:

X = e^{x}

where

X = \frac{1 + \sqrt{5}}{2}

e^{x} = \frac{1 + \sqrt{5}}{2}

Let's apply the ln function on both sides:

\ln (e^{x}) = \ln (\frac{1 + \sqrt{5}}{2})

ⓘ For real

x

\ln (\exp (x)) = x

; wherever

\ln (x)

is defined,

\exp (\ln (x)) = x

\ln (e^{x})

x

▶Solution obtained:

x = \ln (\frac{1 + \sqrt{5}}{2})

4) The numeric value of:

x = \ln (\frac{1 + \sqrt{5}}{2})

is:

x = \ln (\frac{1 + \approx 2.2361}{2})

x = \ln (\frac{\approx 3.2361}{2})

x = \ln (\approx 1.6180)

▷

x = \approx 0.4812