How do you solve #log_8(11-6x)=log_8(1-x)#?

1 Answer
Sep 4, 2016

Answer:

#{O/}#

Explanation:

#log_8(11 - 6x) - log_8(1 - x) = 0#

You need use the apply the rule #log_an - log_am = log_a(n/m)#.

#log_8((11 - 6x)/(1 - x)) = 0#

#(11- 6x)/(1 - x) = 8^0#

#11 - 6x = 1#

#-6x = -10#

#x =5/3#

However, checking in the original equation, you will realize that this solution is extraneous, because if in #log_b(n)#, #n <=0#, the equation will be undefined.

Hence, the solution set is #{O/}#.

Hopefully this helps!