Help with log equations? 10 points guaranteed

Please help! I can't figure solve these out for the life of me...(not a math person)

1.) 2 log6 4 - 1/3 log6 8 = log6 x

2.) log2 x = 1/3 log2 27

Thank you!

Assuming we're dealing with real numbers only:

1.
Assuming all logs in base 6:
2*log(4) - (1/3)*log(8) = log(x)
log(4^2) - log(8^(1/3)) = log(x)
log(16) - log(2) = log(x)
log(16/2) = log(x)
x = 16/2
x = 8

2.
Assuming all logs in base 2:
log(x) = (1/3)*log(27)
log(x) = log(27^(1/3))
log(x) = log(3)
x = 3

2Log6(4) – 1/3Log6(8) = Log6(x) \\ use basic properties of Logarithms

Log6(4²) – Log6(∛8) = Log6(x)

Log6(16/2) = Log6(x)

x = 16/2 = 8

Check this result in the original equation, I did

log_6(16/2) = log_6(x)
x = 8