What is expand log_{a}((2x)/(x-1)) ?

The solution to expand log_{a}((2x)/(x-1)) is log_{a}(2)+log_{a}(x)-log_{a}(x-1)

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expand log_{a}((2x)/(x-1))

expand

lo g_{a} (\frac{2 x}{x - 1})

lo g_{a} (2) + lo g_{a} (x) - lo g_{a} (x - 1)

Solution steps

lo g_{a} (\frac{2 x}{x - 1})

Apply log rule:

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

lo g_{a} (\frac{2 x}{x - 1}) = lo g_{a} (2 x) - lo g_{a} (x - 1)

= lo g_{a} (2 x) - lo g_{a} (x - 1)

Apply log rule:

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

lo g_{a} (2 x) = lo g_{a} (2) + lo g_{a} (x)

= lo g_{a} (2) + lo g_{a} (x) - lo g_{a} (x - 1)

lo g_{10} (x^{3}) y lo g_{10} (x) 3 y

lo g_{3} ((\frac{n ^{3}}{m})^{6})

12 ln (2 x) - tan (4 x)

- \frac{1}{5} ln (- 1) - \frac{1}{5} ln (3)

(\frac{1}{6}) \cdot ln (\frac{11.92}{8.2})

What is expand log_{a}((2x)/(x-1)) ?
The solution to expand log_{a}((2x)/(x-1)) is log_{a}(2)+log_{a}(x)-log_{a}(x-1)