What is expand ,ln((2pi(θ+1))^{-n/2}) ?

The solution to expand ,ln((2pi(θ+1))^{-n/2}) is -(nln(2pi(θ+1)))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand ,ln((2pi(θ+1))^{-n/2})

expand

ln ((2 π (θ + 1))^{- \frac{n}{2}})

- \frac{n ln ( 2 π ( θ + 1 ) )}{2}

Solution steps

ln ((2 π (θ + 1))^{- \frac{n}{2}})

Apply log rule:

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x)

ln ((2 π (θ + 1))^{- \frac{n}{2}}) = - \frac{n}{2} ln (2 π (θ + 1))

= - \frac{n}{2} ln (2 π (θ + 1))

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= - \frac{n ln ( 2 π ( θ + 1 ) )}{2}

ln (e^{4 x - 3})

ln (\frac{0.25}{0.125})

ln (3 (3)^{2})

ln (2 sin (2 x)) + cos (3 x)

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

What is expand ,ln((2pi(θ+1))^{-n/2}) ?
The solution to expand ,ln((2pi(θ+1))^{-n/2}) is -(nln(2pi(θ+1)))/2