3 logx2 + logx18 = 2
3p = 9 (27)q
log2 7 - log2 (11q-2p) = 1
Not clear what you intend the first one to be! Does it have logx2 or logx(something2) ?
Are your last two equations one problem?
If so:
3p = 9·(27)q ---> 3p = 32·(33)q ---> 3p = 32·(33q) ---> 3p = 33q + 2 ---> p = 3q + 2
log27 - log2(11q - 2p) = 1 ---> log2[ 7 / (11q - 2p) ] = 1 ---> 7 / (11q - 2p) = 21
---> 7 = 2( 11q - 2p ) ---> 7 = 22q - 4p
Since p = 3q + 2 ---> 7 = 22q - 4(3q + 2)
---> 7 = 22q - 12q - 8
---> 7 = 10q - 8
---> 15 = 10q ---> q = 1.5
Substituting: p = 6.5