How do you solve # log_6 (x+ 1)+ log_6 (x-4)= 1 #?
1 Answer
Jul 30, 2016
Explanation:
Given:
#log_6(x+1)+log_6(x-4) = 1#
We require:
#6 = 6^1 = 6^(log_6(x+1)+log_6(x-4)) = (x+1)(x-4) = x^2-3x-4#
Subtract
#0 = x^2-3x-10 = (x-5)(x+2)#
So
If
#log_6(x+1)+log_6(x-4) = log_6(6)+log_6(1) = 1+0 = 1#
So
If
#log_6(x+1)+log_6(x-4)#
#= log_6(-1)+log_6(-6)#
#= log_6(1)+(pi i)/ln(6) + log_6(6)+(pi i)/ln(6)#
#= 1+(2pi i)/ln(6) != 1#
So