What is expand log_{6}((16x^2)/(2x-4)) ?

The solution to expand log_{6}((16x^2)/(2x-4)) is 4log_{6}(2)+2log_{6}(x)-log_{6}(2x-4)

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expand log_{6}((16x^2)/(2x-4))

expand

lo g_{6} (\frac{16 x ^{2}}{2 x - 4})

4 lo g_{6} (2) + 2 lo g_{6} (x) - lo g_{6} (2 x - 4)

Solution steps

lo g_{6} (\frac{16 x ^{2}}{2 x - 4})

Apply log rule:

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

lo g_{6} (\frac{16 x ^{2}}{2 x - 4}) = lo g_{6} (16 x^{2}) - lo g_{6} (2 x - 4)

= lo g_{6} (16 x^{2}) - lo g_{6} (2 x - 4)

Apply log rule:

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

lo g_{6} (16 x^{2}) = lo g_{6} (16) + lo g_{6} (x^{2})

= lo g_{6} (16) + lo g_{6} (x^{2}) - lo g_{6} (2 x - 4)

Apply log rule:

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

lo g_{6} (x^{2}) = 2 lo g_{6} (x)

= lo g_{6} (16) + 2 lo g_{6} (x) - lo g_{6} (2 x - 4)

lo g_{6} (16) = 4 lo g_{6} (2)

= 4 lo g_{6} (2) + 2 lo g_{6} (x) - lo g_{6} (2 x - 4)

3 lo g_{10} (2) + lo g_{10} (x) - lo g_{10} (4 x)

lo g_{10} (5 x + 7) - lo g_{10} (x)

ln (9) - ln (2)

lo g_{10} (\frac{n}{2})

ln ∣ 7 - 2 ∣ - \frac{3}{2} ln ∣ 2 (7) - 1 ∣ + 2 ln ∣ 7 + 3 ∣

What is expand log_{6}((16x^2)/(2x-4)) ?
The solution to expand log_{6}((16x^2)/(2x-4)) is 4log_{6}(2)+2log_{6}(x)-log_{6}(2x-4)