How do we solve #logsqrt(x - 8) + 1/2log(2x+ 1) = 1#?

1 Answer
Sep 25, 2016

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!