What is log_{4}(24)-log_{4}(6) ?

The solution to log_{4}(24)-log_{4}(6) is 1

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simplify log_{4}(24)-log_{4}(6)

simplify

lo g_{4} (24) - lo g_{4} (6)

1

Solution steps

lo g_{4} (24) - lo g_{4} (6)

Apply log rule:

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

lo g_{4} (24) - lo g_{4} (6) = lo g_{4} (\frac{24}{6})

= lo g_{4} (\frac{24}{6})

Factor the number:

4 = 2^{2}

= lo g_{2^{2}} (\frac{24}{6})

Apply log rule:

lo g_{a^{b}} (x) = \frac{1}{b} lo g_{a} (x), a > 0

lo g_{2^{2}} (\frac{24}{6}) = \frac{1}{2} lo g_{2} (\frac{24}{6})

= \frac{1}{2} lo g_{2} (\frac{24}{6})

\frac{24}{6} = 2^{2}

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

Apply log rule:

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

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

= \frac{1}{2} \cdot 2

\frac{1}{2} \cdot 2 = 1

= 1

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