Explore a diverse range of topics and get expert answers on IDNLearn.com. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.

Solve the equation. Write the solution set with the exact solutions.

[tex]\[ \ln x + \ln (x - 10) = \ln (9x - 88) \][/tex]

If there is more than one solution, separate the answers with commas.

The exact solution set is [tex]$\square$[/tex]


Sagot :

To solve the equation

[tex]\[ \ln x + \ln (x - 10) = \ln (9x - 88), \][/tex]

we can use properties of logarithms and algebraic manipulation to solve it step-by-step.

### Step-by-Step Solution

1. Combine the logarithmic terms on the left-hand side using the property [tex]\(\ln a + \ln b = \ln (ab)\)[/tex]:

[tex]\[ \ln (x(x - 10)) = \ln (9x - 88). \][/tex]

2. Simplify the argument inside the logarithm:

[tex]\[ \ln \left(x^2 - 10x\right) = \ln (9x - 88). \][/tex]

3. Since the natural logarithms of two expressions are equal, the expressions themselves must be equal:

[tex]\[ x^2 - 10x = 9x - 88. \][/tex]

4. Rearrange the equation to form a standard quadratic equation:

[tex]\[ x^2 - 10x - 9x + 88 = 0 \implies x^2 - 19x + 88 = 0. \][/tex]

5. Solve the quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -19\)[/tex], and [tex]\(c = 88\)[/tex]:

[tex]\[ x = \frac{-(-19) \pm \sqrt{(-19)^2 - 4 \cdot 1 \cdot 88}}{2 \cdot 1}. \][/tex]

6. Calculate the discriminant:

[tex]\[ (-19)^2 - 4 \cdot 1 \cdot 88 = 361 - 352 = 9. \][/tex]

7. Find the solutions:

[tex]\[ x = \frac{19 \pm \sqrt{9}}{2} = \frac{19 \pm 3}{2}. \][/tex]

8. This gives two potential solutions:

[tex]\[ x = \frac{19 + 3}{2} = \frac{22}{2} = 11 \quad \text{and} \quad x = \frac{19 - 3}{2} = \frac{16}{2} = 8. \][/tex]

9. Verify the solutions in the original equation to ensure they do not make the logarithmic terms undefined:

- For [tex]\(x = 11\)[/tex]:
[tex]\[ \ln 11 + \ln (11 - 10) = \ln 11 + \ln 1 = \ln 11 = \ln (9 \cdot 11 - 88) = \ln (99 - 88) = \ln 11. \][/tex]
So, [tex]\(x = 11\)[/tex] is a valid solution.

- For [tex]\(x = 8\)[/tex]:
[tex]\[ \ln 8 + \ln (8 - 10) = \ln 8 + \ln (-2). \][/tex]
The term [tex]\(\ln (-2)\)[/tex] is undefined in the real number system. So, [tex]\(x = 8\)[/tex] is not a valid solution.

### Conclusion

The only valid solution is [tex]\(x = 11\)[/tex]. Thus, the exact solution set is

[tex]\[ \{11\}. \][/tex]