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Log3 - log2^{8} ÷ log3^{9}

From rules of logarithm,

-log(x) = log(1/x)

= Log3 + log(1/2^{8}) ÷ log(3^{9})

= Log3 + log(1/2^{8 }÷ 3^{9})

= Log3 + log(2^{-}^{8 }÷ 3^{9})

From rules of logarithm, log(x ÷ y) = log(x) - log(y)

Thus,

= log3 + log2^{-8} - log3^{9}

= log3 - 8log2 - 9log3

= - 8log2 - 8log3

= 8log(1/2) + 8log(1/3)

From rules of logarithm,

alog(x) + alog(y) = alog(x × y)

= 8log(1/2 × 1/3)

= 8log(1/6)

= log ((1/6)^{8})

it can also be simplified to -log(6^8)

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