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

1) Let's solve in

ℝ

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

ⓘ 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 x)

Since the ln function is strictly increasing,

\ln (a) = \ln (b)

is equivalent to

a = b

x \cdot (x + 1) = 2 x

You can factorize the expression:

x \cdot (x + 1) - 2 x

into:

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

A product is zero if and only if one of its factors is zero: we solve each factor equal to 0

⇥1.1) Let's solve in

ℝ

x = 0

⇥Candidates, pending the domain check:

x = 0

⇥1.2) Let's solve in

ℝ

(x + 1) - 2 = 0

⇥ⓘ Start from

1

and move back

2

on the number line, going below zero:

1 - 2

-1

⇥

x - 1

⇥Candidates, pending the domain check:

x = 1

2) Let's evaluate the domain of

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

For the expression

\ln (x)

to be defined, we must have:

x > 0

For the expression

\ln (x + 1)

to exist, you must check:

x + 1 > 0

For the expression

\ln (2 x)

to exist, you must check:

2 x > 0

Solution obtained:

x > 0

3) Let's solve in

ℝ

x + 1 > 0

ⓘ We subtract

1

from both sides

Solution found:

x > -1

4) Let's solve in

ℝ

2 x > 0

We simplify the expression by

2

⇥Let's solve in

ℝ

x > 0

⇥Solution found:

x > 0

Let's create the table of the different solutions to find the overall solution:

Domain constraints:

x > 0

The solution

x = 0

does not satisfy the domain

x > 0

and is discarded

▶Solution:

x = 1