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To solve the equation [tex]\(\log (x)+\log (16)+\log (x-1)=\log (15)+\log \left(x^2-4\right)\)[/tex], we will use the properties of logarithms and algebraic manipulation. Here is the step-by-step solution:
1. Combine the logarithmic terms on each side using the property [tex]\(\log(a) + \log(b) = \log(ab)\)[/tex]:
[tex]\[ \log(x) + \log(16) + \log(x-1) = \log(15) + \log(x^2 - 4) \][/tex]
This simplifies to:
[tex]\[ \log(16x(x-1)) = \log(15(x^2 - 4)) \][/tex]
2. Set the arguments of the logarithms equal to each other since [tex]\(\log(A) = \log(B) \implies A = B\)[/tex]:
[tex]\[ 16x(x-1) = 15(x^2 - 4) \][/tex]
3. Expand and simplify the equation to form a single polynomial equation:
[tex]\[ 16x^2 - 16x = 15(x^2 - 4) \][/tex]
[tex]\[ 16x^2 - 16x = 15x^2 - 60 \][/tex]
4. Move all terms to one side to set the equation to zero:
[tex]\[ 16x^2 - 16x - 15x^2 + 60 = 0 \][/tex]
[tex]\[ x^2 - 16x + 60 = 0 \][/tex]
5. Solve the quadratic equation [tex]\(x^2 - 16x + 60 = 0\)[/tex]. We can do this using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -16\)[/tex], and [tex]\(c = 60\)[/tex]:
[tex]\[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 60}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{256 - 240}}{2} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{16 \pm 4}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{20}{2} = 10 \quad \text{and} \quad x = \frac{12}{2} = 6 \][/tex]
6. Verify the solutions by checking if they satisfy the original equation and the domain constraints (logarithm arguments must be positive):
- For [tex]\(x = 10\)[/tex]:
[tex]\[ x = 10 \quad \Rightarrow \quad \log(10) + \log(16) + \log(9) \][/tex]
[tex]\[ = \log(16 \cdot 10 \cdot 9) = \log(1440) \][/tex]
Compare with:
[tex]\[ \log(15) + \log(96) = \log(15 \cdot 96) = \log(1440) \][/tex]
Both sides are equal, so [tex]\(x = 10\)[/tex] is a valid solution.
- For [tex]\(x = 6\)[/tex]:
[tex]\[ x = 6 \quad \Rightarrow \quad \log(6) + \log(16) + \log(5) \][/tex]
[tex]\[ = \log(16 \cdot 6 \cdot 5) = \log(480) \][/tex]
Compare with:
[tex]\[ \log(15) + \log(32) = \log(15 \cdot 32) = \log(480) \][/tex]
Both sides are equal, so [tex]\(x = 6\)[/tex] is a valid solution.
Therefore, the solutions to the equation [tex]\(\log (x)+\log (16)+\log (x-1)=\log (15)+\log \left(x^2-4\right)\)[/tex] are:
[tex]\[ x = 6 \quad \text{and} \quad x = 10 \][/tex]
1. Combine the logarithmic terms on each side using the property [tex]\(\log(a) + \log(b) = \log(ab)\)[/tex]:
[tex]\[ \log(x) + \log(16) + \log(x-1) = \log(15) + \log(x^2 - 4) \][/tex]
This simplifies to:
[tex]\[ \log(16x(x-1)) = \log(15(x^2 - 4)) \][/tex]
2. Set the arguments of the logarithms equal to each other since [tex]\(\log(A) = \log(B) \implies A = B\)[/tex]:
[tex]\[ 16x(x-1) = 15(x^2 - 4) \][/tex]
3. Expand and simplify the equation to form a single polynomial equation:
[tex]\[ 16x^2 - 16x = 15(x^2 - 4) \][/tex]
[tex]\[ 16x^2 - 16x = 15x^2 - 60 \][/tex]
4. Move all terms to one side to set the equation to zero:
[tex]\[ 16x^2 - 16x - 15x^2 + 60 = 0 \][/tex]
[tex]\[ x^2 - 16x + 60 = 0 \][/tex]
5. Solve the quadratic equation [tex]\(x^2 - 16x + 60 = 0\)[/tex]. We can do this using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -16\)[/tex], and [tex]\(c = 60\)[/tex]:
[tex]\[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 60}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{256 - 240}}{2} \][/tex]
[tex]\[ x = \frac{16 \pm \sqrt{16}}{2} \][/tex]
[tex]\[ x = \frac{16 \pm 4}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{20}{2} = 10 \quad \text{and} \quad x = \frac{12}{2} = 6 \][/tex]
6. Verify the solutions by checking if they satisfy the original equation and the domain constraints (logarithm arguments must be positive):
- For [tex]\(x = 10\)[/tex]:
[tex]\[ x = 10 \quad \Rightarrow \quad \log(10) + \log(16) + \log(9) \][/tex]
[tex]\[ = \log(16 \cdot 10 \cdot 9) = \log(1440) \][/tex]
Compare with:
[tex]\[ \log(15) + \log(96) = \log(15 \cdot 96) = \log(1440) \][/tex]
Both sides are equal, so [tex]\(x = 10\)[/tex] is a valid solution.
- For [tex]\(x = 6\)[/tex]:
[tex]\[ x = 6 \quad \Rightarrow \quad \log(6) + \log(16) + \log(5) \][/tex]
[tex]\[ = \log(16 \cdot 6 \cdot 5) = \log(480) \][/tex]
Compare with:
[tex]\[ \log(15) + \log(32) = \log(15 \cdot 32) = \log(480) \][/tex]
Both sides are equal, so [tex]\(x = 6\)[/tex] is a valid solution.
Therefore, the solutions to the equation [tex]\(\log (x)+\log (16)+\log (x-1)=\log (15)+\log \left(x^2-4\right)\)[/tex] are:
[tex]\[ x = 6 \quad \text{and} \quad x = 10 \][/tex]
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