What is log_{b}(28)-log_{b}(7) ?

The solution to log_{b}(28)-log_{b}(7) is 2log_{b}(2)

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simplify log_{b}(28)-log_{b}(7)

simplify

lo g_{b} (28) - lo g_{b} (7)

2 lo g_{b} (2)

Solution steps

lo g_{b} (28) - lo g_{b} (7)

Apply log rule:

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

lo g_{b} (28) - lo g_{b} (7) = lo g_{b} (\frac{28}{7})

= lo g_{b} (\frac{28}{7})

Divide the numbers:

\frac{28}{7} = 4

= lo g_{b} (4)

Factor the number:

4 = 2^{2}

= lo g_{b} (2^{2})

Apply log rule:

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

lo g_{b} (2^{2}) = 2 lo g_{b} (2)

= 2 lo g_{b} (2)

lo g_{10} (\frac{z ^{5}}{y x ^{3}})

2 lo g_{10} (3 x)

lo g_{10} (\frac{243}{32}) + lo g_{10} (\frac{32}{243})

lo g_{10} (\frac{1}{10 x})

ln (6) - ln (5)

What is log_{b}(28)-log_{b}(7) ?
The solution to log_{b}(28)-log_{b}(7) is 2log_{b}(2)