Question

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

Answer 1

Given

a² + b² = 7ab

Adding both sides 2ab we get

a² + b² + 2ab = 7ab + 2ab

(a + b)² = 9ab

`("a" + "b")^2/9` = ab

`(("a" + "b")/3)^2` = ab

Taking square root on both sides

`("a" + "b")/3 = sqrt("ab")`

`("a" + "b")/3 = ("ab")^(1/2)`

Taking logarithm on both sides

`log (("a" + "b")/3) = log ("ab")^(1/2)`

`log (("a" + "b")/3) = 1/2 log("ab")`

`log (("a" + "b")/3) = 1/2 (log "a" + log "b")`