Logarithm relation --Help "thats the word"

[solver]

Originally posted in logarithm relation thread,could someone close that thread please. Thanks

Can any explain help me solve this question:
a^{2}+ b^2=7ab Prove that log{x}{a+b}over{3}={1}over{2}(log{x}a+log{x}b

 

[ShortRubiksFreak]

what have you tried so far??

 

[solver]

nothing really, i am totally stumped

 

[K073]

a^2 + b^2 = 7ab
(a+b)^2 - 2ab = 7ab
(a+b)^2 = 9ab
log{x} (a+b)^2 = log{x} 9ab
2 log{x} (a+b) = log{x} 9 + log{x} ab
log{x} (a+b) = 1/2 (log{x} 9 + log{x} ab)
log{x} (a+b) = log{x} 3 + 1/2(log{x} a + log{x} b)
log{x} (a+b) - log{x} 3 = 1/2(log{x} a + log{x} b)
log{x} [(a+b) / 3] = 1/2(log{x} a + log{x} b)

I did almost full step for you

 

[solver]

How did u get the - 2ab?

 

[nitrocan]

log{x}{a+b}over{3}={1}over{2}(log{x}a+log{x}b)
=logx((a+b)/3) = 1/2(logx(a*b)
=logx((a+b)/3) = logx(sqrt(a*b))
=logx((a+b)/3) - logx(sqrt(a*b)) = 0
logx((a+b)/(3 * sqrt(a*b))) = 0
(a+b)/(3*sqrt(a*b) = 1
a+b = 3*sqrt(a*b)
a^2 + b^2 + 2ab = 9ab
a^2 + b^2 = 7ab

 

[solver]

@K073 i know where the -2ab comes from. Thanks for the help

 

[solver]

Also thanks nitrocan for your help.