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Question
Answer the following:
Show that `7log (15/16) + 6log(8/3) + 5log (2/5) + log(32/25)` = log 3
Solution
L.H.S. = `7log (15/16) + 6log(8/3) + 5log (2/5) + log(32/25)`
= `log(15/16)^7 + log(8/3)^6 + log(2/5)^5 + log(32/25)`
= `log((3 xx 5)/2^4)^7 + log(2^3/3)^6 + log(2/5)^5 + log(2^5/5^2)`
= `log((3^7 xx 5^7)/2^28) + log(2^18/3^6) + log(2^5/5^5) + log(2^5/5^2)`
= `log[(3^7 xx 5^7)/2^28 xx 2^18/3^6 xx 2^5/5^5 xx 2^5/5^2]`
= log 3
= R.H.S.
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