Sum of solutions of the equation \[ \log_{x-3}(6x^2 + 28x + 30) = 5 - 2\log_{x-10}(x^2 + 6x + 9) \] are:

Question

If the domain of \[ f(x)=\log_{(10x^2-17x+7)}\,(18x^2-11x+1) \] is $(-\infty,a)\cup(b,c)\cup(d,\infty)-\{e\}$, then find $90(a+b+c+d+e)$.

Answer 1

To solve the equation $\log_{x-3}(6x^2 + 28x + 30) = 5 - 2\log_{x-10}(x^2 + 6x + 9)$, we follow these steps:

Consider the constraints for the logarithmic functions to be defined:
- $x - 3 > 0$ and $x - 10 > 0$, leading to $x > 10$ because $x - 10 > x - 3$.
- $6x^2 + 28x + 30 > 0$ and $x^2 + 6x + 9 > 0$.
Let's first solve for $(x^2 + 6x + 9) = 0$ to check when the logarithm becomes undefined:
- The equation $x^2 + 6x + 9 = 0$ is a perfect square: $(x+3)^2 = 0$. This implies $x = -3$, which is not within our valid range.
Rewrite the equation using properties of logarithms:
- The equation can be transformed using:
  $2\log_{x-10}(x^2 + 6x + 9) = \log_{(x-10)^2}((x+3)^2)$
- Original equation becomes:
  $\log_{x-3}(6x^2 + 28x + 30) = \log_{(x-10)^2}((x+3)^2) \Rightarrow 6x^2 + 28x + 30 = (x+3)^2$
Now, solve for: $6x^2 + 28x + 30 = (x+3)^2$:
- Expand and simplify: $(x+3)^2 = x^2 + 6x + 9$
- Equating and simplifying:
  $6x^2 + 28x + 30 = x^2 + 6x + 9 \Rightarrow 5x^2 + 22x + 21 = 0$
Determine the roots using the quadratic formula: $ax^2 + bx + c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
- $a = 5, b = 22, c = 21$
- Discriminant: $b^2 - 4ac = 22^2 - 4 \cdot 5 \cdot 21 = 484 - 420 = 64$
- Roots: $x = \frac{-22 \pm \sqrt{64}}{10} = \frac{-22 \pm 8}{10}$
- Solving:
  $x = \frac{-22 + 8}{10} = -1.4$ and $x = \frac{-22 - 8}{10} = -3$
Since both solutions $x = -1.4$ and $x = -3$ do not satisfy the condition $x > 10$, they are not valid for the equation.
Thus, there are no valid solutions. Therefore, the sum of solutions is $0$.

The correct answer is therefore 0.

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