Solve the equation log base 3 of (2x^2 + 13x + 23) divided by log base 3 of (x + 5) equals 1, and find the sum of its roots.

Step-by-step solution

1. 1

   Write down the domain first, because the denominator hides two separate conditions. Both logarithm arguments must be positive, and the denominator must not be zero:

   $x+5>0,\qquad \log_3(x+5)\neq 0\ \Rightarrow\ x+5\neq 1,\qquad 2x^2+13x+23>0.$

   The first two give $x>-5$ and $x\neq -4$.

2. 2

   Check the quadratic argument separately - it turns out to be free. Its discriminant is

   $D=13^2-4\cdot 2\cdot 23=169-184=-15<0,$

   and the leading coefficient is positive, so $2x^2+13x+23>0$ for every real $x$. That condition adds nothing, and the domain is simply $x>-5$, $x\neq-4$.

3. 3

   Recognise the quotient of logs as a change of base. Since $\dfrac{\log_3 A}{\log_3 B}=\log_B A$, the equation says

   $\log_{x+5}\left(2x^2+13x+23\right)=1.$

   A logarithm equals $1$ exactly when the argument equals the base, which is why we can drop the logarithms entirely instead of exponentiating twice.

4. 4

   Solve the resulting quadratic. Setting argument equal to base:

   $2x^2+13x+23=x+5\ \Longrightarrow\ 2x^2+12x+18=0\ \Longrightarrow\ x^2+6x+9=0\ \Longrightarrow\ (x+3)^2=0,$

   so $x=-3$ is a repeated root of the quadratic.

5. 5

   Check the candidate and add up the roots. $x=-3$ satisfies $x>-5$ and $x\neq-4$, and substituting back gives $\log_3 2/\log_3 2=1$, so it is genuine. Although the quadratic has $x=-3$ twice, the original equation has exactly one root, so the sum of its roots is

   $-3.$