logsub3 12to the power of 5/6 Xto the power of 9

log3(125/6)9

use rule of exponents (BA)C = BAC

log3125/6·9

log31215/2

now use the rule of logs:  logBAC = C·logBA

15/2·log312

Now write 12 in terms of the base 3, i.e., 12 = (3·4)

15/2·log3(3·4)

Nou use the rule of logs:   logB(AC) = (logBA + logBC)

15/2(log33 + log34)

Finally use the rule of logs logBB = 1

15/2(1 + log34)

Edwin
AnlytcPhil@aol.com