Get expert advice and community support for your questions on IDNLearn.com. Find reliable solutions to your questions quickly and accurately with help from our dedicated community of experts.

Order the steps to solve the equation [tex]\log \left(x^2-15\right)=\log (2x)[/tex] from 1 to 5.

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 \text{ or } x + 3 = 0[/tex]
5. Potential solutions are [tex]x = -3[/tex] and [tex]x = 5[/tex]


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]