log_4(x^2+3x) - log_4(x-5) = 1log4(x2+3x)−log4(x−5)=1
Using the identity log_ab -log_ac = log_a(b/c)logab−logac=loga(bc), we can simplify the expression to
log_4((x^2+3x)/(x-5)) = 1log4(x2+3xx−5)=1
This yields to
(x^2+3x)/(x-5) = 4^1x2+3xx−5=41
This is because, if log_ab = clogab=c, then b=a^cb=ac
Now the equation can be solved for values of xx
=>x^2+3x = 4^1(x-5)⇒x2+3x=41(x−5)
=> x^2+3x = 4x -20⇒x2+3x=4x−20
=> x^2 + 3x -4x +20 = 0⇒x2+3x−4x+20=0
=> x^2-x+20=0⇒x2−x+20=0
The roots of this equation are imaginary.
They can be found using the formula x = (-b +- sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a
Here, a=1a=1, b=-1b=−1 and c=20c=20
Therefore, x= (1+- sqrt((-1)^2-4*1*20))/(2*1)x=1±√(−1)2−4⋅1⋅202⋅1
=>x = (1+sqrt(79)i)/2, (1-sqrt(79)i)/2⇒x=1+√79i2,1−√79i2
