Use this property of logarithms:
[tex]\log_b\left(\dfrac{x}{y}\right)=\log_b x-\log_b y[/tex]
Your equation transforms into:
[tex]log_4\left(\dfrac{x^2+3x}{x+5}\right)=1[/tex]
Now, you have to apply the definition of a logarithm to express the equation in exponential form:
[tex]\dfrac{x^2+3x}{x+5} = 4^1[/tex]
In case you don't remember, this is the definition of a logarithm:
[tex]\log_b x = y \iff b^y = x[/tex]
The log is the exponent (y) you have to raise the base (b) to in order to get the power (x).
Finally, solve the rational equation:
[tex]\dfrac{x^2+3x}{x+5} =4 \iff x^2+3x=4(x+5) [/tex]
[tex]\iff x^2-x-20=0 [/tex]
[tex]\iff x=\dfrac{1\pm\sqrt{(-1)^2-4\cdot(-20)}}{2}= \left \{ {{5} \atop {-4}} \right. [/tex]
The correct answer is d.
