What is log_{32}(1/64) ?

The solution to log_{32}(1/64) is -6/5

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simplify log_{32}(1/64)

simplify

lo g_{32} (\frac{1}{64})

- \frac{6}{5}

Decimal Notation

- 1.2

Solution steps

lo g_{32} (\frac{1}{64})

Factor the number:

32 = 2^{5}

= lo g_{2^{5}} (\frac{1}{64})

Apply log rule:

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

lo g_{2^{5}} (\frac{1}{64}) = \frac{1}{5} lo g_{2} (\frac{1}{64})

= \frac{1}{5} lo g_{2} (\frac{1}{64})

\frac{1}{64} = 2^{- 6}

= \frac{1}{5} lo g_{2} (2^{- 6})

Apply log rule:

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

lo g_{2} (2^{- 6}) = - 6

= \frac{1}{5} (- 6)

\frac{1}{5} (- 6) = - \frac{6}{5}

= - \frac{6}{5}

- 1 + - 81 - 16 - - 1

\frac{x ^{2} - 3 x y - 10 y ^{2}}{5 y ^{2} + 4 x y - x ^{2}}

\frac{\frac{( x ^{2} - 1 )}{24}}{\frac{( x + 1 ) ^{2}}{36}}

(3 a - b)^{2}