Question 87641
Starting with {{{ log((a+b)/3) = (1/2)(loga + logb) }}} i shall re-write it as {{{ 2log((a+b)/3) = (loga + logb) }}}.

I am going to prove this is correct instead by looking at the Left hand side {{{ 2log((a+b)/3) }}} and showing that it equals the right hand side {{{ (loga + logb) }}}.


So,
{{{ 2log((a+b)/3) }}}
{{{ log(((a+b)/3))^2 }}} --> one of laws of Logs
{{{ log((a^2+2ab+b^2)/9) }}} --> multiply out brackets


now we are told that {{{ a^2+b^2 = 7ab }}} so sub that into what we have and we get:
{{{ log((a^2+b^2+2ab)/9) }}} re-arrange order so you can see the squared terms together
{{{ log((7ab+2ab)/9) }}} --> sub in the other equality
{{{ log((9ab)/9) }}} --> add together like terms
{{{ log(ab) }}} --> cancel out the 9's
{{{ log(a) + log(b) }}} use one of the Laws of Logs


which is what i set out to prove it equalled.


cheers
jon