What is [ln(s^2-1)-ln(s^2+1)] ?

The solution to [ln(s^2-1)-ln(s^2+1)] is ln((s^2-1)/(s^2+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify [ln(s^2-1)-ln(s^2+1)]

simplify

[ln (s^{2} - 1) - ln (s^{2} + 1)]

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

Solution steps

[ln (s^{2} - 1) - ln (s^{2} + 1)]

Apply log rule:

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

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

= [ln (\frac{s ^{2} - 1}{s ^{2} + 1})]

Remove parentheses:

(a) = a

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

5.37 \cdot 1 0^{- 7} + lo g_{10} (\frac{0.0056}{0.0266})

e ln (2 e)

\frac{3}{2} lo g_{10} (15)

ln (\frac{x ^{2}}{x - 1})

- 0.76 - 0.02958 \cdot lo g_{10} (\frac{1}{0.3})

What is [ln(s^2-1)-ln(s^2+1)] ?
The solution to [ln(s^2-1)-ln(s^2+1)] is ln((s^2-1)/(s^2+1))