Question

Answer the following:

If a² + b² = 7ab, show that, `log(("a" + "b")/3) = 1/2 log "a" + 1/2 log "b"`

Answer 1

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"`