What is log_{4}(7/19)+log_{4}(19/7) ?

The solution to log_{4}(7/19)+log_{4}(19/7) is 0

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simplify log_{4}(7/19)+log_{4}(19/7)

simplify

lo g_{4} (\frac{7}{19}) + lo g_{4} (\frac{19}{7})

0

Solution steps

lo g_{4} (\frac{7}{19}) + lo g_{4} (\frac{19}{7})

Apply log rule:

lo g_{a} (x) + lo g_{a} (y) = lo g_{a} (x y)

lo g_{4} (\frac{7}{19}) + lo g_{4} (\frac{19}{7}) = lo g_{4} (\frac{7}{19} \cdot \frac{19}{7})

= lo g_{4} (\frac{7}{19} \cdot \frac{19}{7})

Factor the number:

4 = 2^{2}

= lo g_{2^{2}} (\frac{7}{19} \cdot \frac{19}{7})

Apply log rule:

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

lo g_{2^{2}} (\frac{7}{19} \cdot \frac{19}{7}) = \frac{1}{2} lo g_{2} (\frac{7}{19} \cdot \frac{19}{7})

= \frac{1}{2} lo g_{2} (\frac{7}{19} \cdot \frac{19}{7})

Cross cancel:

\frac{a}{b} \cdot \frac{b}{a} = 1

\frac{7}{19} \cdot \frac{19}{7} = 1

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

Apply log rule:

lo g_{a} (1) = 0

lo g_{2} (1) = 0

= \frac{1}{2} \cdot 0

Apply rule:

a \cdot 0 = 0

= 0

7 lo g_{6} (x - 2) + 2 lo g_{6} (3 x)

lo g_{10} (0.0001) - lo g_{10} (0.1)

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

ln (\frac{( d - r ) ^{2}}{r ^{2}})

\frac{1}{2} (lo g_{5} (s) + lo g_{5} (t))

What is log_{4}(7/19)+log_{4}(19/7) ?
The solution to log_{4}(7/19)+log_{4}(19/7) is 0