What is log_{4}(1/6)+log_{1/3}(81) ?

The solution to log_{4}(1/6)+log_{1/3}(81) is -4-log_{4}(6)

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simplify log_{4}(1/6)+log_{1/3}(81)

simplify

lo g_{4} (\frac{1}{6}) + lo g_{\frac{1}{3}} (81)

- 4 - lo g_{4} (6)

Decimal Notation

- 5.29248 \dots

Solution steps

lo g_{4} (\frac{1}{6}) + lo g_{\frac{1}{3}} (81)

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

lo g_{4} (\frac{1}{6}) = lo g_{4} (1) - lo g_{4} (6)

= lo g_{4} (1) - lo g_{4} (6) + lo g_{\frac{1}{3}} (81)

Apply log rule:

lo g_{a} (1) = 0

= 0 - lo g_{4} (6) + lo g_{\frac{1}{3}} (81)

0 - lo g_{4} (6) + lo g_{\frac{1}{3}} (81) = - lo g_{4} (6) + lo g_{\frac{1}{3}} (81)

= lo g_{\frac{1}{3}} (81) - lo g_{4} (6)

lo g_{\frac{1}{3}} (81) = - 4

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

lo g_{3} (\frac{x}{x - 2})

lo g_{x} (18 a^{5}) - 2 lo g_{x} (3 a)

lo g_{3} (81 x^{3} 1 + x^{2})

lo g_{4} (x y^{4} z^{4})

lo g_{4} (x) - lo g_{4} (2) + lo g_{4} (3)

What is log_{4}(1/6)+log_{1/3}(81) ?
The solution to log_{4}(1/6)+log_{1/3}(81) is -4-log_{4}(6)