Question

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

Answer 1

`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`