What is log_{10}(θ^x(1-θ)^{1-x}) ?

The solution to log_{10}(θ^x(1-θ)^{1-x}) is xlog_{10}(θ)+(1-x)log_{10}(1-θ)

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simplify log_{10}(θ^x(1-θ)^{1-x})

simplify

lo g_{10} (θ^{x} (1 - θ)^{1 - x})

x lo g_{10} (θ) + (1 - x) lo g_{10} (1 - θ)

Solution steps

lo g_{10} (θ^{x} (1 - θ)^{1 - x})

Apply log rule:

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

lo g_{10} (θ^{x} (1 - θ)^{1 - x}) = lo g_{10} (θ^{x}) + lo g_{10} ((1 - θ)^{1 - x})

= lo g_{10} (θ^{x}) + lo g_{10} ((- θ + 1)^{- x + 1})

Apply log rule:

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

lo g_{10} (θ^{x}) = x lo g_{10} (θ)

= x lo g_{10} (θ) + lo g_{10} ((1 - θ)^{1 - x})

Apply log rule:

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

lo g_{10} ((1 - θ)^{1 - x}) = (1 - x) lo g_{10} (1 - θ)

= x lo g_{10} (θ) + (1 - x) lo g_{10} (1 - θ)

- lo g_{10} (8.1 \cdot 1 0^{- 9})

2 ln ∣ x + 2 ∣ - ln ∣ x + 3 ∣ + C

lo g_{5} (y) - 4 lo g_{5} (z)

ln (12.5) + ln (e^{1.6 t})

ln (x^{2}) + ln (2)

What is log_{10}(θ^x(1-θ)^{1-x}) ?
The solution to log_{10}(θ^x(1-θ)^{1-x}) is xlog_{10}(θ)+(1-x)log_{10}(1-θ)