What is log_{9}(11/17)+log_{9}(17/11) ?

The solution to log_{9}(11/17)+log_{9}(17/11) is 0

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simplify log_{9}(11/17)+log_{9}(17/11)

simplify

lo g_{9} (\frac{11}{17}) + lo g_{9} (\frac{17}{11})

0

Solution steps

lo g_{9} (\frac{11}{17}) + lo g_{9} (\frac{17}{11})

Apply log rule:

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

lo g_{9} (\frac{11}{17}) + lo g_{9} (\frac{17}{11}) = lo g_{9} (\frac{11}{17} \cdot \frac{17}{11})

= lo g_{9} (\frac{11}{17} \cdot \frac{17}{11})

Factor the number:

9 = 3^{2}

= lo g_{3^{2}} (\frac{11}{17} \cdot \frac{17}{11})

Apply log rule:

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

lo g_{3^{2}} (\frac{11}{17} \cdot \frac{17}{11}) = \frac{1}{2} lo g_{3} (\frac{11}{17} \cdot \frac{17}{11})

= \frac{1}{2} lo g_{3} (\frac{11}{17} \cdot \frac{17}{11})

Cross cancel:

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

\frac{11}{17} \cdot \frac{17}{11} = 1

= \frac{1}{2} lo g_{3} (1)

Apply log rule:

lo g_{a} (1) = 0

lo g_{3} (1) = 0

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

Apply rule:

a \cdot 0 = 0

= 0

ln (\frac{17}{7})

lo g_{b} (x yz)

6 ln (4) - 2 ln (7) + 6 ln (12) - 2 ln (5)

lo g_{10} (\frac{y ^{4}}{100 x})

ln (8 e^{x})

What is log_{9}(11/17)+log_{9}(17/11) ?
The solution to log_{9}(11/17)+log_{9}(17/11) is 0