What is log_{32}(5^{log_{5}(2)}) ?

The solution to log_{32}(5^{log_{5}(2)}) is 1/5

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simplify log_{32}(5^{log_{5}(2)})

simplify

lo g_{32} (5^{l o g_{5} (2)})

\frac{1}{5}

Decimal Notation

0.2

Solution steps

lo g_{32} (5^{l o g_{5} (2)})

Factor the number:

32 = 2^{5}

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

Apply log rule:

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

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

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

Apply log rule:

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

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

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

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{1}{5} \cdot \frac{ln ( 2 )}{ln ( 5 )} lo g_{2} (5)

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

Cross cancel:

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

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

= \frac{1}{5} \cdot 1

Multiply the numbers:

\frac{1}{5} \cdot 1 = \frac{1}{5}

= \frac{1}{5}

7 x

(2 x - 3 y)^{5}

(11 + i) (4 - 4 i)

- 1 + (- 4)

\frac{1}{x} + x

What is log_{32}(5^{log_{5}(2)}) ?
The solution to log_{32}(5^{log_{5}(2)}) is 1/5