How do you solve log_4 (x^2 + 3x) - log_4(x - 5) =1log4(x2+3x)log4(x5)=1?

1 Answer

x=(1+sqrt(79)i)/2, (1-sqrt(79)i)/2x=1+79i2,179i2

Explanation:

log_4(x^2+3x) - log_4(x-5) = 1log4(x2+3x)log4(x5)=1
Using the identity log_ab -log_ac = log_a(b/c)logablogac=loga(bc), we can simplify the expression to
log_4((x^2+3x)/(x-5)) = 1log4(x2+3xx5)=1
This yields to
(x^2+3x)/(x-5) = 4^1x2+3xx5=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(x5)
=> x^2+3x = 4x -20x2+3x=4x20
=> x^2 + 3x -4x +20 = 0x2+3x4x+20=0
=> x^2-x+20=0x2x+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±b24ac2a
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)2412021
=>x = (1+sqrt(79)i)/2, (1-sqrt(79)i)/2x=1+79i2,179i2