What is log_{10}(t^x(1-t)^{1-x}) ?

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

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simplify log_{10}(t^x(1-t)^{1-x})

simplify

lo g_{10} (t^{x} (1 - t)^{1 - x})

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

Solution steps

lo g_{10} (t^{x} (1 - t)^{1 - x})

Apply log rule:

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

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

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

Apply log rule:

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

lo g_{10} (t^{x}) = x lo g_{10} (t)

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

Apply log rule:

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

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

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

0.48 - \frac{0.0592}{6} lo g_{10} (\frac{1.5}{0.5})

ln (\frac{19}{44})

- lo g_{10} (3.98 \cdot 1 0^{- 9})

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