Solve the simultaneous equations 4^x=6 and 2(2^x)=3(2^y)
from 4^x = 6
2^(2x) = 6
(2^x)^2 = 6
from 2(2^x) = 3(2^y)
2^x = (3/2)(2^y)
(2^x)^2 = (9/4)(2^(2y))
(9/4)(2^(2y)) = 6
2^(2y) = 8/3
take logs of both sides and use log rules
2y(log2) = log8 - log3
2y = (log8-log3)/log2 = 1.04150....
y = .7075....
from 4^x = 6
x = log6/log4 = 1.2924...
or, could have found x first from 4^x = 6
then sub that into the 2nd equation.
since 4^x = 6, x = 2.2925
2*2^x = 3*2^y
2*2^(x-y) = 3
4*2^(x-y) = 6
2^(x-y+2) = 6
and 2^(2x) = 6
so 2x = x-y+2
x+y = 2
so y = 1.7075