The sum of all possible real values of x for which $\log_{x-3}(x^2-9)=\log_{x-3}(x+1)+2 , is
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Official Correct Answer: C. Okay, so I have this logarithm problem to solve. It's from the CAT exam, and We find the sum of all possible real values of x that satisfy the equation:
$$
\log_{x-3}(x^2 - 9) = \log_{x-3}(x + 1) + 2
$$
Hmm, logarithms can be tricky, especially with different bases and variables inside. Let me try to break this down step by step.
First, for a logarithm to be defined, the base must be positive and not equal to 1, and the argument must be positive. So, I need to consider the conditions for the base and the arguments here.
The base of the logarithm is \( x - 3 \). So, \( x - 3 > 0 \) implies \( x > 3 \), and \( x - 3 \neq 1 \) implies \( x \neq 4 \). So, x must be greater than 3 but not equal to 4.
Next, the arguments inside the logarithms must be positive. So:
1. \( x^2 - 9 > 0 \) implies \( (x - 3)(x + 3) > 0 \). Since x > 3, both \( x - 3 \) and \( x + 3 \) are positive, so this is satisfied.
2. \( x + 1 > 0 \) implies \( x > -1 \). But since we already have x > 3, this is automatically satisfied.
So, the domain for x is \( x > 3 \) and \( x \neq 4 \).
Now, looking at the equation:
$$
\log_{x-3}(x^2 - 9) = \log_{x-3}(x + 1) + 2
$$
I can use the property of logarithms that says \( \log_b a = \log_b c + d \) implies \( a = c \cdot b^d \). So, applying this here:
Let me rewrite the equation:
$$
\log_{x-3}(x^2 - 9) - \log_{x-3}(x + 1) = 2
$$
Using the logarithm subtraction rule, which is \( \log_b a - \log_b c = \log_b \left( \frac{a}{c} \right) \), I can combine the logs:
$$
\log_{x-3}\left( \frac{x^2 - 9}{x + 1} \right) = 2
$$
Simplify the fraction inside the log:
\( x^2 - 9 \) factors into \( (x - 3)(x + 3) \), so:
$$
\frac{(x - 3)(x + 3)}{x + 1} = \frac{(x - 3)(x + 3)}{x + 1}
$$
So, the equation becomes:
$$
\log_{x-3}\left( \frac{(x - 3)(x + 3)}{x + 1} \right) = 2
$$
Since \( x - 3 \) is the base, I can rewrite the logarithmic equation in exponential form. Remember that \( \log_b a = c \) implies \( a = b^c \). So:
$$
\frac{(x - 3)(x + 3)}{x + 1} = (x - 3)^2
$$
Simplify this equation:
Multiply both sides by \( x + 1 \) to eliminate the denominator:
$$
(x - 3)(x + 3) = (x - 3)^2 (x + 1)
$$
Expanding both sides.
Left side:
\( (x - 3)(x + 3) = x^2 - 9 \)
Right side:
First, expand \( (x - 3)^2 \):
\( (x - 3)^2 = x^2 - 6x + 9 \)
Then multiply by \( (x + 1) \):
\( (x^2 - 6x + 9)(x + 1) \)
Let me multiply term by term:
- \( x^2 \cdot x = x^3 \)
- \( x^2 \cdot 1 = x^2 \)
- \( -6x \cdot x = -6x^2 \)
- \( -6x \cdot 1 = -6x \)
- \( 9 \cdot x = 9x \)
- \( 9 \cdot 1 = 9 \)
Combine like terms:
- \( x^3 \)
- \( x^2 - 6x^2 = -5x^2 \)
- \( -6x + 9x = 3x \)
- \( +9 \)
So, the right side becomes:
\( x^3 - 5x^2 + 3x + 9 \)
Now, the equation is:
Left side: \( x^2 - 9 \)
Right side: \( x^3 - 5x^2 + 3x + 9 \)
Bring all terms to one side:
\( x^2 - 9 - x^3 + 5x^2 - 3x - 9 = 0 \)
Combine like terms:
- \( -x^3 \)
- \( x^2 + 5x^2 = 6x^2 \)
- \( -3x \)
- \( -9 - 9 = -18 \)
So, the equation becomes:
\( -x^3 + 6x^2 - 3x - 18 = 0 \)
Multiply both sides by -1 to make it positive:
\( x^3 - 6x^2 + 3x + 18 = 0 \)
Now, I need to solve this cubic equation. Factoring it.
Looking for rational roots using Rational Root Theorem. Possible roots are factors of 18 over factors of 1: ±1, ±2, ±3, ±6, ±9, ±18.
Let me test x=3:
\( 27 - 54 + 9 + 18 = 0 \). 27 -54 is -27, +9 is -18, +18 is 0. So, x=3 is a root.
So, (x - 3) is a factor. Let's perform polynomial division or factor it out.
Divide \( x^3 - 6x^2 + 3x + 18 \) by (x - 3).
Using synthetic division:
3 | 1 -6 3 18
3 -9 -18
1 -3 -6 0
So, the cubic factors as (x - 3)(x^2 -