What is 1/2 (log_{6}(r-1)-log_{6}(r)) ?

The solution to 1/2 (log_{6}(r-1)-log_{6}(r)) is (log_{6}((r-1)/r))/(2)

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simplify 1/2 (log_{6}(r-1)-log_{6}(r))

simplify

\frac{1}{2} (lo g_{6} (r - 1) - lo g_{6} (r))

\frac{lo g _{6} ( \frac{r - 1}{r} )}{2}

Solution steps

\frac{1}{2} (lo g_{6} (r - 1) - lo g_{6} (r))

Apply log rule:

lo g_{a} (x) - lo g_{a} (y) = lo g_{a} (\frac{x}{y})

lo g_{6} (r - 1) - lo g_{6} (r) = lo g_{6} (\frac{r - 1}{r})

= \frac{1}{2} lo g_{6} (\frac{r - 1}{r})

Apply the fraction rule:

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

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

Apply rule:

1 \cdot a = a

1 \cdot lo g_{6} (\frac{r - 1}{r}) = lo g_{6} (\frac{r - 1}{r})

= \frac{lo g _{6} ( \frac{r - 1}{r} )}{2}

lo g_{10} (\frac{a ^{4} b ^{3}}{c ^{2}})

7 lo g_{3} (u) + 2 lo g_{3} (v)

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

3 \cdot lo g_{10} (3) - 2 \cdot lo g_{10} (6)

lo g_{3} (5 x)

What is 1/2 (log_{6}(r-1)-log_{6}(r)) ?
The solution to 1/2 (log_{6}(r-1)-log_{6}(r)) is (log_{6}((r-1)/r))/(2)