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

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

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

expand

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

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

Solution steps

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

Apply log rule:

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

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

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

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

- \frac{6 ln ( \frac{9}{10} )}{ln ( 2 )}

ln (5) - ln (2) + 39

lo g_{10} (3) - lo g_{10} (7)

- lo g_{10} (1.34 \cdot 1 0^{- 5})

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