What is ln(θ^y(1-θ^{n-y})) ?

The solution to ln(θ^y(1-θ^{n-y})) is yln(θ)+ln(1-θ^{n-y})

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simplify ln(θ^y(1-θ^{n-y}))

simplify

ln (θ^{y} (1 - θ^{n - y}))

y ln (θ) + ln (1 - θ^{n - y})

Solution steps

ln (θ^{y} (1 - θ^{n - y}))

Apply log rule:

lo g_{c} (ab) = lo g_{c} (a) + lo g_{c} (b)

ln (θ^{y} (1 - θ^{n - y})) = ln (θ^{y}) + ln (1 - θ^{n - y})

= ln (θ^{y}) + ln (- θ^{- y + n} + 1)

Apply log rule:

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

ln (θ^{y}) = y ln (θ)

= y ln (θ) + ln (1 - θ^{n - y})

lo g_{10} (5 x^{4} \cdot (1 - x)^{12 - 4})

ln (A e^{- \frac{B}{RT}})

a - ln (3 a)

ln \frac{7}{5} + \frac{3 x}{5} - ln - \frac{3}{5} + \frac{3}{5} x

lo g_{6} (2) - lo g_{6} (4) + lo g_{6} (3)

What is ln(θ^y(1-θ^{n-y})) ?
The solution to ln(θ^y(1-θ^{n-y})) is yln(θ)+ln(1-θ^{n-y})