Solve 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

   Recognise the change-of-base pattern. A quotient of two logarithms with the same base is a single logarithm whose base is the argument of the denominator:

   $\frac{\log_3 A}{\log_3 B}=\log_B A.$

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

2. 2

   Write down the domain restrictions before solving. For that logarithm to exist we need

   $x+5>0,\qquad x+5\neq 1,\qquad 2x^{2}+13x+23>0,$

   that is $x>-5$ and $x\neq -4$. These conditions do the real work at the end.

3. 3

   Convert to algebraic form. $\log_B A=1$ means $A=B$:

   $2x^{2}+13x+23=x+5.$

4. 4

   Solve the quadratic.

   $2x^{2}+12x+18=0\ \Longrightarrow\ x^{2}+6x+9=0\ \Longrightarrow\ (x+3)^{2}=0\ \Longrightarrow\ x=-3,$

   a repeated root, so there is only one distinct solution candidate.

5. 5

   Check it against the domain. At $x=-3$ the base is $x+5=2$, which is positive and not $1$; the argument is $2(9)+13(-3)+23=18-39+23=2>0$. Both sides become $\log_2 2=1$, so $x=-3$ is genuine.

6. 6

   Report the sum of the roots. The equation has the single root $x=-3$, so

   $\text{sum of roots}=-3.$

   Note the trap: the quadratic $x^2+6x+9$ has coefficient sum $-6$ by Vieta, but one of its two (coincident) roots is not a second solution — the repeated root is one number, and Vieta's sum would have been the wrong answer had the roots been distinct with one rejected by the domain.