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

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

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simplify 2log_{a}(x)-log_{a}(x-1)-log_{a}(x-2)

simplify

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

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

Solution steps

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

Apply log rule:

a lo g_{c} (b) = lo g_{c} (b^{a})

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply the fraction rule:

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

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

ln (\frac{r}{8 s})

lo g_{10} (6) + 2 lo g_{10} (x) + 4 lo g_{10} (y)

lo g_{6} ((3 \cdot 11)^{6})

\frac{4 ln ( \frac{54}{7} )}{ln ( 2 )}

ln (\frac{1}{3})

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