What is expand log_{4}(1/16 x) ?

The solution to expand log_{4}(1/16 x) is log_{4}(x)-2

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expand log_{4}(1/16 x)

expand

lo g_{4} (\frac{1}{16} x)

lo g_{4} (x) - 2

Solution steps

lo g_{4} (\frac{1}{16} x)

Apply log rule:

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

lo g_{4} (\frac{1}{16} x) = lo g_{4} (\frac{1}{16}) + lo g_{4} (x)

= lo g_{4} (\frac{1}{16}) + lo g_{4} (x)

Apply log rule:

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

lo g_{4} (\frac{1}{16}) = lo g_{4} (1) - lo g_{4} (16)

= lo g_{4} (1) - lo g_{4} (16) + lo g_{4} (x)

Apply log rule:

lo g_{a} (1) = 0

= 0 - lo g_{4} (16) + lo g_{4} (x)

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

= lo g_{4} (x) - lo g_{4} (16)

lo g_{4} (16) = 2

= lo g_{4} (x) - 2

2 lo g_{10} (a) + 3 lo g_{10} (b)

lo g_{8} (x z^{3})

lo g_{10} (z) - lo g_{10} (13)

lo g_{6} (\frac{g}{36})

lo g_{7} (\frac{x ^{3} y}{18})

What is expand log_{4}(1/16 x) ?
The solution to expand log_{4}(1/16 x) is log_{4}(x)-2