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Sagot :
Sure, let's order the steps to solve the equation [tex]\(\log (x^2 - 15) = \log (2x)\)[/tex].
1. Remove the logarithms by equating the arguments:
[tex]\[ x^2 - 15 = 2x \][/tex]
2. Bring all terms to one side to form a quadratic equation:
[tex]\[ x^2 - 2x - 15 = 0 \][/tex]
3. Factorize the quadratic equation:
[tex]\[ (x - 5)(x + 3) = 0 \][/tex]
4. Solve for the potential solutions:
[tex]\[ x - 5 = 0 \quad \text{or} \quad x + 3 = 0 \\ x = 5 \quad \text{or} \quad x = -3 \][/tex]
5. Verify potential solutions satisfy the original equation:
[tex]\[ \text{Potential solutions are } -3 \text{ and } 5 \][/tex]
Now, putting these steps in the correct order:
1. [tex]\[ x^2 - 15 = 2x \][/tex]
2. [tex]\[ x^2 - 2x - 15 = 0 \][/tex]
3. [tex]\[ (x - 5)(x + 3) = 0 \][/tex]
4. [tex]\[ x - 5 = 0 \quad \text{or} \quad x + 3 = 0 \\ (x-5)(x+3)=0 \][/tex]
5. [tex]\[ \text{Potential solutions are } -3 \text{ and } 5 \][/tex]
1. Remove the logarithms by equating the arguments:
[tex]\[ x^2 - 15 = 2x \][/tex]
2. Bring all terms to one side to form a quadratic equation:
[tex]\[ x^2 - 2x - 15 = 0 \][/tex]
3. Factorize the quadratic equation:
[tex]\[ (x - 5)(x + 3) = 0 \][/tex]
4. Solve for the potential solutions:
[tex]\[ x - 5 = 0 \quad \text{or} \quad x + 3 = 0 \\ x = 5 \quad \text{or} \quad x = -3 \][/tex]
5. Verify potential solutions satisfy the original equation:
[tex]\[ \text{Potential solutions are } -3 \text{ and } 5 \][/tex]
Now, putting these steps in the correct order:
1. [tex]\[ x^2 - 15 = 2x \][/tex]
2. [tex]\[ x^2 - 2x - 15 = 0 \][/tex]
3. [tex]\[ (x - 5)(x + 3) = 0 \][/tex]
4. [tex]\[ x - 5 = 0 \quad \text{or} \quad x + 3 = 0 \\ (x-5)(x+3)=0 \][/tex]
5. [tex]\[ \text{Potential solutions are } -3 \text{ and } 5 \][/tex]
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