If a² + b² = 7ab, show that Log [1/3 (a+b)] = 1/2 (Log a + Log b)...
So... (a+b)² = a² + 2ab + b² then group them into "a² + b² +2ab" and replace "a² +b²" with 7ab since the "if = 7ab" to make 7ab + 2ab... which equals 9ab...
Then... (a+b)² = 9ab... take the square root of both sides to get (a+b) = 3√ab
After that use the log properties.. then uh...
Log (a+b) = Log 3 + Log a½ + Log b½... then subtract Log 3 on both sides to get Log (a+b) - Log 3 = Log a½ + Log b½...
By using the log properties "Logь (x/y) = Logьx - Logьy" and "Logь(xy) = Logьx + Logьy" and "Logьx^r = r Logьx" I'll end up with...
Log (a+b)/3 = ½ (Log a + Log b)
[1/3 (a+b)] = (a+b)/3....
So... uh... is this it? Or do I have to solve it? It says show... and I think I showed it but I'm not sure...
You did great, only a bit off. It wants you to prove from
=Log[1/3 (a+b)]
and change it into
=1/2[log a +log b]