Let a,b be positive real numbers that satisfy
2 + log2(2a) = 3 + log4(2b) + log6(a+b)
Find (a+b)/ab
+118608 is that
\(log_2(2a)\qquad or \qquad log_{10}(2*2a)\)
I assume it is the first one?
+128407 We can write
2 + log (2a) / log 2 = 3 + log (2b) / log 4 + log (a + b) / log 6
2 + (log 2 + log a) / (log 2) = 3 + (log 2) (log b) / log (2)^2 + log (a + b) / log 6
2 + 1 + log a / log 2 = 3 + (log 2) (log b) / (2 log 2) + log (a + b) / log 6
log a / log 2 = log b /2 + log (a + b) / log 6
( 1/2) (log a - log b) = log ( a +b) / log 6
log a - log b = 2 log (a + b)/log 6
log 6 * log (a/ b) = 2 log ( a + b)
log 6 / 2 = log (a + b) / (log (a/b))
Note that this will be true when a = 4 and b = 2
Check :
2 + log2 8 = 3 + log 4 4 + log 6 6
2 + 3 = 3 + 1 + 1
5 = 5
So (a + b) /(ab) = 6 / 8 = 3 / 4
