What is expand log_{4}(((x+2)^2)/(2x)) ?

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

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expand log_{4}(((x+2)^2)/(2x))

expand

lo g_{4} (\frac{( x + 2 ) ^{2}}{2 x})

2 lo g_{4} (x + 2) - \frac{1}{2} - lo g_{4} (x)

Solution steps

lo g_{4} (\frac{( x + 2 ) ^{2}}{2 x})

Apply log rule:

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

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

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

lo g_{4} (2) = \frac{1}{2}

= 2 lo g_{4} (x + 2) - (lo g_{4} (x) + \frac{1}{2})

- (\frac{1}{2} + lo g_{4} (x)) : - \frac{1}{2} - lo g_{4} (x)

= 2 lo g_{4} (x + 2) - \frac{1}{2} - lo g_{4} (x)

2 ln (6 x)

lo g_{10} (\frac{x ^{7} z ^{9}}{y})

2 (lo g_{10} (3) - lo g_{10} (5))

ln (16 e)

lo g_{10} (a) \frac{1}{4} - lo g_{10} (b) \frac{1}{2}

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