What is log_{n}(n+1)<log_{n-1}(n) ?

The solution to log_{n}(n+1)<log_{n-1}(n) is log_{n}(n+1)-log_{n-1}(n)<0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{n}(n+1)<log_{n-1}(n)

lo g_{n} (n + 1) < lo g_{n - 1} (n)

lo g_{n} (n + 1) - lo g_{n - 1} (n) < 0

Solution steps

lo g_{n} (n + 1) < lo g_{n - 1} (n)

Move

lo g_{n - 1} (n)

to the left side

lo g_{n} (n + 1) - lo g_{n - 1} (n) < 0

Apply log rules

lo g_{n} (n + 1) - lo g_{n - 1} (n) < 0

lo g_{0.2} (x + 5) > 0

\frac{x}{( lo g _{\frac{1}{2}} ( 2 x + 3 ) )} > 0

x \cdot ln (x^{3}) < 0

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

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

What is log_{n}(n+1)<log_{n-1}(n) ?
The solution to log_{n}(n+1)<log_{n-1}(n) is log_{n}(n+1)-log_{n-1}(n)<0