What is 1/7 (log_{10}(x)-log_{10}(x+4)) ?

The solution to 1/7 (log_{10}(x)-log_{10}(x+4)) is (log_{10}(x/(x+4)))/7

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simplify 1/7 (log_{10}(x)-log_{10}(x+4))

simplify

\frac{1}{7} (lo g_{10} (x) - lo g_{10} (x + 4))

\frac{lo g _{10} ( \frac{x}{x + 4} )}{7}

Solution steps

\frac{1}{7} (lo g_{10} (x) - lo g_{10} (x + 4))

Apply log rule:

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

lo g_{10} (x) - lo g_{10} (x + 4) = lo g_{10} (\frac{x}{x + 4})

= \frac{1}{7} lo g_{10} (\frac{x}{x + 4})

Apply the fraction rule:

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

= \frac{1 \cdot lo g _{10} ( \frac{x}{x + 4} )}{7}

Apply rule:

1 \cdot a = a

1 \cdot lo g_{10} (\frac{x}{x + 4}) = lo g_{10} (\frac{x}{x + 4})

= \frac{lo g _{10} ( \frac{x}{x + 4} )}{7}

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

ln (x^{5} (1 - x)^{3})

- \frac{8}{9} lo g_{2} (\frac{8}{9}) - \frac{1}{9} lo g_{2} (\frac{1}{9})

\frac{1}{2} lo g_{b} (x) + 3 lo g_{b} (y) - 2 lo g_{b} (x)

lo g_{8} (10) - lo g_{8} (2)

What is 1/7 (log_{10}(x)-log_{10}(x+4)) ?
The solution to 1/7 (log_{10}(x)-log_{10}(x+4)) is (log_{10}(x/(x+4)))/7