What is the answer to log_{e}(x+1)+log_{e}(x)=0 ?

The answer to log_{e}(x+1)+log_{e}(x)=0 is x=(-1+sqrt(5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{e}(x+1)+log_{e}(x)=0

lo g_{e} (x + 1) + lo g_{e} (x) = 0

x = \frac{- 1 + 5}{2}

Decimal Notation

x = 0.61803 \dots

Solution steps

lo g_{e} (x + 1) + lo g_{e} (x) = 0

Apply log rules

(x + 1) x = 1

Solve

(x + 1) x = 1 : x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

x = \frac{- 1 + 5}{2}, x = \frac{- 1 - 5}{2}

Verify Solutions:

x = \frac{- 1 + 5}{2}

True

, x = \frac{- 1 - 5}{2}

False

The solution is

x = \frac{- 1 + 5}{2}

lo g_{x - 4} (x) = 3

ln (x) = - ln (x)

lo g_{10} (x) - lo g_{10} (x - 2) = 1

lo g_{27} (a^{2}) = 18

lo g_{3} (4 - 3^{x - 1} - 1) = 2 x - 1

What is the answer to log_{e}(x+1)+log_{e}(x)=0 ?
The answer to log_{e}(x+1)+log_{e}(x)=0 is x=(-1+sqrt(5))/2