What is ln((3z-1)/(z^2+1)) ?

The solution to ln((3z-1)/(z^2+1)) is ln(3z-1)-ln(z^2+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ln((3z-1)/(z^2+1))

simplify

ln (\frac{3 z - 1}{z ^{2} + 1})

ln (3 z - 1) - ln (z^{2} + 1)

Solution steps

ln (\frac{3 z - 1}{z ^{2} + 1})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{3 z - 1}{z ^{2} + 1}) = ln (3 z - 1) - ln (z^{2} + 1)

= ln (3 z - 1) - ln (z^{2} + 1)

ln (\frac{1}{10}) + ln (e^{3})

- 2 ln (\frac{2 - 3}{2})

ln (\frac{x + 2 y}{y}) - ln (y)

- ln (x) - ln (x - 1)

ln (8) + 5 ln (x) + 4 ln (x^{2} + 3)

What is ln((3z-1)/(z^2+1)) ?
The solution to ln((3z-1)/(z^2+1)) is ln(3z-1)-ln(z^2+1)