What is the answer to 2log_{e}(x)+log_{e}(x)=0 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2log_{e}(x)+log_{e}(x)=0

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

x = 1

Solution steps

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

Add similar elements:

2 lo g_{e} (x) + lo g_{e} (x) = 3 lo g_{e} (x)

3 lo g_{e} (x) = 0

Divide both sides by

3

lo g_{e} (x) = 0

Apply log rules

x = 1

Verify Solutions:

x = 1

True

The solution is

x = 1

lo g_{5} (P) = 10

ln (x^{3}) = 7 ln (x)

3 ln (4) = 2 ln (x)

lo g_{3} (2 x - 7) = lo g_{3} (13 - 3 x)

ln (x - 2) + 1 = 0

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