How do you solve #log_4 (x^2 + 3x) - log_4(x - 5) =1#?

1 Answer

Answer:

#x=(1+sqrt(79)i)/2, (1-sqrt(79)i)/2#

Explanation:

#log_4(x^2+3x) - log_4(x-5) = 1#
Using the identity #log_ab -log_ac = log_a(b/c)#, we can simplify the expression to
#log_4((x^2+3x)/(x-5)) = 1#
This yields to
#(x^2+3x)/(x-5) = 4^1#
This is because, if #log_ab = c#, then #b=a^c#
Now the equation can be solved for values of #x#
#=>x^2+3x = 4^1(x-5)#
#=> x^2+3x = 4x -20#
#=> x^2 + 3x -4x +20 = 0#
#=> x^2-x+20=0#
The roots of this equation are imaginary.
They can be found using the formula #x = (-b +- sqrt(b^2-4ac))/(2a)#
Here, #a=1#, #b=-1# and #c=20#
Therefore, #x= (1+- sqrt((-1)^2-4*1*20))/(2*1)#
#=>x = (1+sqrt(79)i)/2, (1-sqrt(79)i)/2#