What is log_{16}(7^{log_{7}(4)}) ?

The solution to log_{16}(7^{log_{7}(4)}) is 1/2

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simplify log_{16}(7^{log_{7}(4)})

simplify

lo g_{16} (7^{l o g_{7} (4)})

\frac{1}{2}

Solution steps

lo g_{16} (7^{l o g_{7} (4)})

Factor the number:

16 = 2^{4}

= lo g_{2^{4}} (7^{l o g_{7} (4)})

Apply log rule:

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

lo g_{2^{4}} (7^{l o g_{7} (4)}) = \frac{1}{4} lo g_{2} (7^{l o g_{7} (4)})

= \frac{1}{4} lo g_{2} (7^{l o g_{7} (4)})

lo g_{7} (4) = 2 lo g_{7} (2)

= \frac{1}{4} lo g_{2} (7^{2 l o g_{7} (2)})

Apply log rule:

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

lo g_{2} (7^{2 l o g_{7} (2)}) = lo g_{7} (4) lo g_{2} (7)

= \frac{1}{4} lo g_{7} (4) lo g_{2} (7)

Apply log rule:

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

lo g_{7} (4) = \frac{ln ( 4 )}{ln ( 7 )}

= \frac{1}{4} \cdot \frac{ln ( 4 )}{ln ( 7 )} lo g_{2} (7)

Apply log rule:

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

lo g_{2} (7) = \frac{ln ( 7 )}{ln ( 2 )}

= \frac{1}{4} \cdot \frac{ln ( 4 )}{ln ( 7 )} \cdot \frac{ln ( 7 )}{ln ( 2 )}

Cross-cancel common factor:

ln (7)

= \frac{1}{4} \cdot \frac{ln ( 4 )}{ln ( 2 )}

ln (4) = 2 ln (2)

= \frac{1}{4} \cdot \frac{2 ln ( 2 )}{ln ( 2 )}

Cancel the common factor:

ln (2)

= \frac{1}{4} \cdot 2

\frac{1}{4} \cdot 2 = \frac{1}{2}

= \frac{1}{2}

- sin (x) sin (x)

x^{2} + \frac{1}{x}

\frac{1}{2} + \frac{4}{6} + \frac{1}{3}

\frac{( \frac{2}{5} ) ^{- 3}}{( \frac{3}{7} ) ^{- 3}}

What is log_{16}(7^{log_{7}(4)}) ?
The solution to log_{16}(7^{log_{7}(4)}) is 1/2