How do you solve #log_2((-9x)/(2x^2-1)) = 1# ?

Question

941 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-30T09:04:40

#x=(-9-sqrt(123))/8#

#log_2((-9x)/(2x^2-1)) = log_2(-9x)-log_2(2x^2-1) = 1 = log_2 2#

Since #log_2(x)# is a one-one function (as a Real function), we require:

#(-9x)/(2x^2-1) = 2#

Multiplying both sides by #(2x^2-1)# we get:

#-9x = 4x^2-2#

Add #9x# to both sides and transpose to get:

#4x^2+9x-2 = 0#

#x = (-9+-sqrt(9^2-(4*4*-2)))/(2*4)#

#=(-9+-sqrt(81+32))/8#

#=(-9+-sqrt(123))/8#

We can discard #(-9+sqrt(123))/8 > 0#, since it results in #-9x < 0#, so the Real logarithm is not defined.

That leaves #x=(-9-sqrt(123))/8# as the only solution of the original equation.