What is 1/2 (log_{6}(k^9)-log_{6}(k^3)) ?

The solution to 1/2 (log_{6}(k^9)-log_{6}(k^3)) is 3log_{6}(k)

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simplify 1/2 (log_{6}(k^9)-log_{6}(k^3))

simplify

\frac{1}{2} (lo g_{6} (k^{9}) - lo g_{6} (k^{3}))

3 lo g_{6} (k)

Solution steps

\frac{1}{2} (lo g_{6} (k^{9}) - lo g_{6} (k^{3}))

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), assuming x > 0

lo g_{6} (k^{9}) = 9 lo g_{6} (k)

= \frac{1}{2} (9 lo g_{6} (k) - lo g_{6} (k^{3}))

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x), assuming x > 0

lo g_{6} (k^{3}) = 3 lo g_{6} (k)

= \frac{1}{2} (9 lo g_{6} (k) - 3 lo g_{6} (k))

Add similar elements:

9 lo g_{6} (k) - 3 lo g_{6} (k) = 6 lo g_{6} (k)

= \frac{1}{2} \cdot 6 lo g_{6} (k)

Apply the fraction rule:

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

= \frac{1 \cdot 6 lo g _{6} ( k )}{2}

Multiply the numbers:

1 \cdot 6 = 6

= \frac{6 lo g _{6} ( k )}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3 lo g_{6} (k)

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

f (x) = 5 lo g_{10} (x^{3}) - 2 lo g_{10} (x^{4})

lo g_{b} (\frac{p ^{3} q ^{6}}{m ^{4} b ^{7}})

lo g_{10} (3 x + 9) + lo g_{10} (x + 8)

lo g_{b} (x) + lo g_{b} (x + 6) - lo g_{b} (24)

What is 1/2 (log_{6}(k^9)-log_{6}(k^3)) ?
The solution to 1/2 (log_{6}(k^9)-log_{6}(k^3)) is 3log_{6}(k)