The value of ( 1 27 ) 2 - ( log 5 16 2 log 5 9 ) = 0 . p q , then p + q =

Question

The value of ${(\frac{1}{\sqrt{27}})}^{2 - (\frac{\log_{5} 16}{2 \log_{5} 9})} = 0 . p q$ , then $p + q =$ ________ .

Smart Achievers · Accepted Answer

(1)

$2 - \frac{\log_{5} 16}{2 \log_{5} 9} = 2 - \frac{\log_{5} 16}{\log_{5} 81} = 2 - \log_{81} 16$

$= 2 - \log_{3} 2 = \log_{3} 9 - \log_{3} 2 = \log_{3} \frac{9}{2}$

${(\frac{1}{\sqrt{27}})}^{\log_{3} \frac{9}{2}} = {(3^{- \frac{3}{2}})}^{\log_{3} \frac{9}{2}} = 3^{\log_{3}}$ ${(\frac{9}{2})}^{- \frac{3}{2}}$

${(\frac{9}{2})}^{- \frac{3}{2}} = {(\frac{2}{9})}^{\frac{3}{2}} = \frac{2 \sqrt{2}}{27} = 0.10 = 0 . p q$

$∴$ $p + q = 1 + 0 = 1$

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