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Question
Answer the following:
If a2 + b2 = 7ab, show that, `log(("a" + "b")/3) = 1/2 log "a" + 1/2 log "b"`
Solution
a2 + b2 = 7ab
∴ a2 + b2 + 2ab = 7ab + 2ab
∴ (a + b)2 = 9ab
∴ `("a" + "b")^2/9` = ab
∴ `(("a" + "b")/3)^2` = ab
∴ `log(("a" + "b")/3)^2` = log (ab)
∴ `2log(("a" + "b")/3)` = log a + log b
∴ `log(("a" + "b")/3) = 1/2 log"a" + 1/2log"b"`
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