math logs question?

Question

http://img504.imageshack.us/img504/1048/mathbq1.jpg

The answer is 2x + 3y.
How do I arrive at the answer?
A step by step solution will be appreciated.

Answer 1

log (base 8) 5 = x ... ---> 8^x = 5
log (base 4) 7 = y ... ---> 4^y = 7
8^x = 5
4^y = 7
2^2y = 7
2^3x = 5
2^2y * 2^3x = 5 * 7
2^(2y+3x) = 35
log (base 2) 35 = 2y + 3x

Answer 2

log_8 5 = x --> 8^x = 5 --> 2^(3x) = 5 --> log _2 5 = 3x
log_ 4 7 =y --> 4^y =7 -> 2^(2y) = 7 --> log_2 7 = 2y
log_2 5+log_2 7 = log_2 35 = 3x+2y

Answer 3

log 5 / log 8 = log 5 / log (2^3) = log 5 /3log2 = x
so log 5 /log 2 = 3x

log 7/log 4 = log 7 /log (2^2) = log 7 /2log2 = y
log 7 /log 2 = 2y

since you want to express log2 (35) which equals to log 35 / log 2
log 35 /log 2 = (log 7+log5)/log2 = (log7/log2)+(log5/log2)
so log35 /log 2 = 2y + 3x