another pre cal (logs)
posted by Deb on .
write as a single log
log(underscore3)(3x6)[log(underscore3)(x^24)+log(underscore3)81]
thanks

according to your basic laws of logs your expression is
log_{3} ((3x6)(x^2  4)/81) 
I misread you + and  signs
should be
log_{3} (81(3x6)/(x^2  4)) 
which reduces to
log_{3} 243/(x2) , x > 2 
Thank you