Question #8fd03

1 Answer
Nov 1, 2017

#x=1/3#

Explanation:

#log_m27+log_mx^4=log_mx#

using the law

#log_aX+log_aY=log_aXY#

we have

#log_m27x^4=log_mx#

#=>27x^4=x#

#=>27x^4-x=0#

#x(27x^3-1)=0#

either #x=0#

but #""log_aX" "#is not defined for #X<=0#
#:. x=0 " is not a solution"#

#or 27x^3-1=0#

#=>x^3=1/27#

#x=1/3#