What is ln((1-xy)/(1+x^2y^2))+(1-xy)/(1+x^2y^2) ?

The solution to ln((1-xy)/(1+x^2y^2))+(1-xy)/(1+x^2y^2) is ln(1-xy)-ln(1+x^2y^2)+(1-xy)/(1+x^2y^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

simplify ln((1-xy)/(1+x^2y^2))+(1-xy)/(1+x^2y^2)

simplify

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

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

Solution steps

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

Apply log rule:

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

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

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

lo g_{8} (\frac{1}{625})

ln (\frac{9.96087}{0.46087})

ln (- 1.2 \cdot 1 0^{4})

ln (\frac{sin ( x )}{y})

ln (11) - ln (8)

What is ln((1-xy)/(1+x^2y^2))+(1-xy)/(1+x^2y^2) ?
The solution to ln((1-xy)/(1+x^2y^2))+(1-xy)/(1+x^2y^2) is ln(1-xy)-ln(1+x^2y^2)+(1-xy)/(1+x^2y^2)