ln(x)+ln(x+1)=ln(2) — step-by-step solution

1) Let's solve in

ℝ

\ln (x) + \ln (x + 1) = \ln (2)

ⓘ For strictly positive arguments, a sum of logarithms is the logarithm of the product, a difference that of the quotient, ln(a) + ln(b) = ln(a*b):

\ln (x) + \ln (x + 1)

\ln (x \cdot (x + 1))

\ln (x \cdot (x + 1)) = \ln (2)

On the logarithm's domain, ln is strictly increasing:

\ln (x \cdot (x + 1)) = \ln (2)

is equivalent to

x \cdot (x + 1) = 2

ⓘ We subtract

2

from both sides

x \cdot (x + 1) - 2 = 0

ⓘ Expand by multiplying each term inside the parentheses:

x \cdot (x + 1)

x^{2} + x

x^{2} + x - 2 = 0

Given the quadratic form

a x^{2} + b x + c

, you can calculate the discriminant:

Δ = b^{2} - 4 a c

. Here

a = 1

b = 1

and

c = -2

Δ = 1 - 4 \cdot 1 \cdot (-2)

Δ = 1 - 4 \cdot (-2)

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

- 4 \cdot (-2)

8

Δ = 1 + 8

ⓘ Start from

1

and move forward

8

on the number line:

1 + 8

9

Δ = 9

Since

Δ > 0

, the equation has two roots:

The quadratic formula gives

x_{1}

\frac{- b - \sqrt{Δ}}{2 a}

\frac{- 1 - \sqrt{9}}{2 \cdot 1}

x_{1} = \frac{- 1 - \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{1} = \frac{- 1 - 3}{2}

ⓘ The numerator and the denominator are computed before the division:

- 1 - 3

-4

x_{1} = - \frac{4}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

- \frac{4}{2}

-2

x_{1} = -2

The quadratic formula gives

x_{2}

\frac{- b + \sqrt{Δ}}{2 a}

\frac{- 1 + \sqrt{9}}{2 \cdot 1}

x_{2} = \frac{- 1 + \sqrt{9}}{2}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{9}

3

;

3^{2}

9

x_{2} = \frac{- 1 + 3}{2}

ⓘ The numerator and the denominator are computed before the division:

- 1 + 3

2

x_{2} = \frac{2}{2}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

2

, their greatest common divisor:

\frac{2}{2}

1

x_{2} = 1

Candidates, pending the domain check:

x = {-2, 1}

2) Let's evaluate the domain of

\ln (x) + \ln (x + 1) = \ln (2)

For the expression

\ln (x)

to exist, you must check:

x > 0

For the expression

\ln (x + 1)

to be defined, we must have:

x + 1 > 0

Solution obtained:

x > 0

3) Let's solve in

ℝ

x + 1 > 0

ⓘ We subtract

1

from both sides

Solution found:

x > -1

Intersecting

x > 0

and

x > -1

gives

x > 0

Domain constraints:

x > 0

The solution

x = -2

does not satisfy the domain

x > 0

and is discarded

▶Solution:

x = 1