We have to start by simplifying using the rule #alogn = logn^a#.
#logsqrt(x - 8) + logsqrt(2x + 1) = 1#
Now, use the rule #logn + logm = log(n xx m)#
#log(sqrt(x - 8) xx sqrt(2x + 1)) = 1#
#log(sqrt(2x^2 - 15x - 8)) = 1#
#sqrt(2x^2 - 15x - 8) = 10^1#
#(sqrt(2x^2 - 15x - 8))^2 = (10^1)^2#
#2x^2 - 15x - 8 = 100#
#2x^2 - 15x - 108 = 0#
#2x^2 - 24x + 9x - 108 = 0#
#2x(x - 12) + 9(x - 12) = 0#
#(2x+ 9)(x - 12) = 0#
#x = -9/2 and 12#
However, #x = -9/2# is extraneous since it renders the square root negative, which is undefined in the real number system. Hence, the solution set is #{12}#.
Hopefully this helps!
