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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 -3 and 5


Sagot :

To order the steps to solve the equation [tex]\(\log \left(x^2-15\right)=\log(2x)\)[/tex] from 1 to 5, follow these steps:

1. Start with the given equation [tex]\(\log \left(x^2-15\right) = \log (2 x)\)[/tex].
Since [tex]\(\log(a) = \log(b)\)[/tex] implies [tex]\(a = b\)[/tex], we can set the arguments of the logarithms equal to each other:
[tex]\[ x^2 - 15 = 2x \][/tex]

2. Rearrange the equation to standard quadratic form:
[tex]\[ x^2 - 2x - 15 = 0 \][/tex]

3. Factorize the quadratic equation:
[tex]\[ (x - 5)(x + 3) = 0 \][/tex]

4. Solve for [tex]\(x\)[/tex] by setting each factor equal to zero:
[tex]\[ x - 5 = 0 \text { or } x + 3 = 0 \][/tex]

5. Find the potential solutions:
[tex]\[ x = 5 \text { or } x = -3 \][/tex]

Thus, the ordered steps are:

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 -3 and 5