What is log_{5}(2^{-log_{2}(5)}) ?

The solution to log_{5}(2^{-log_{2}(5)}) is -1

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simplify log_{5}(2^{-log_{2}(5)})

simplify

lo g_{5} (2^{- l o g_{2} (5)})

- 1

Solution steps

lo g_{5} (2^{- l o g_{2} (5)})

Apply log rule:

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

lo g_{5} (2^{- l o g_{2} (5)}) = - lo g_{2} (5) lo g_{5} (2)

= - lo g_{2} (5) lo g_{5} (2)

Apply log rule:

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

lo g_{2} (5) = \frac{ln ( 5 )}{ln ( 2 )}

= - \frac{ln ( 5 )}{ln ( 2 )} lo g_{5} (2)

Apply log rule:

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

lo g_{5} (2) = \frac{ln ( 2 )}{ln ( 5 )}

= - \frac{ln ( 5 )}{ln ( 2 )} \cdot \frac{ln ( 2 )}{ln ( 5 )}

Cross cancel:

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

\frac{ln ( 5 )}{ln ( 2 )} \cdot \frac{ln ( 2 )}{ln ( 5 )} = 1

= - 1

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\frac{2 - x}{x - 2}

(\frac{5}{2} c + \frac{8}{4}) (\frac{6}{3} c + \frac{4}{8})

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What is log_{5}(2^{-log_{2}(5)}) ?
The solution to log_{5}(2^{-log_{2}(5)}) is -1