What is log_{2}(x+1)+log_{2}(x-1) ?

The solution to log_{2}(x+1)+log_{2}(x-1) is log_{2}((x+1)(x-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify log_{2}(x+1)+log_{2}(x-1)

simplify

lo g_{2} (x + 1) + lo g_{2} (x - 1)

lo g_{2} ((x + 1) (x - 1))

Solution steps

lo g_{2} (x + 1) + lo g_{2} (x - 1)

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

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

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

\frac{1}{3} lo g_{2} (5)

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

lo g_{10} (z y^{6})

2 lo g_{10} (3 x) - lo g_{10} (x + 1) - 3 lo g_{10} (x)

ln (x (x + 5))

What is log_{2}(x+1)+log_{2}(x-1) ?
The solution to log_{2}(x+1)+log_{2}(x-1) is log_{2}((x+1)(x-1))