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Question
Prove that `log 2 + 16log 16/15 + 12log 25/24 + 7log 81/80` = 1
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Solution
`log 2 + 16log 16/15 + 12log 25/24 + 7log 81/80`
= `log 2 + log (16/15)^16 + log (25/24)^12 + log (81/80)^7`
=`log 2 + log (2^4/(3 xx 5))^16 + log (5^2/(2^3 xx 3))^12 + log (3^4/(2^4 xx 5))^7`
= `log [2 xx (2^4/(3 xx 5))^16 xx (5^2/(2^3 xx 3))^12 xx (3^4/(2^4 xx 5))^7]`
= `log [2 xx 2^64/(3^16 xx 5^16) xx 5^24/(2^36 xx 3^12) xx 3^28/(2^28 xx 5^7)]`
= `log [(2^65 xx 5^24 xx 3^28)/(3^28 xx 5^23 xx 2^64)]`
= log 2 × 5
= log1010
= 1
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