To solve the equation \((\ln x)^2 + 2\ln x - 15 = 0\), let's proceed with the following steps:

### Step 1: Substitute \( y = \ln x \)
This substitution simplifies the equation to:
\[
y^2 + 2y - 15 = 0
\]

### Step 2: Solve the quadratic equation
The quadratic equation can be solved using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 1\), \(b = 2\), and \(c = -15\). Plugging in these values:
\[
y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-15)}}{2(1)}
\]
\[
y = \frac{-2 \pm \sqrt{4 + 60}}{2}
\]
\[
y = \frac{-2 \pm \sqrt{64}}{2}
\]
\[
y = \frac{-2 \pm 8}{2}
\]

This gives two solutions for \(y\):
\[
y = \frac{-2 + 8}{2} = \frac{6}{2} = 3
\]
\[
y = \frac{-2 - 8}{2} = \frac{-10}{2} = -5
\]

### Step 3: Back-substitute \(y = \ln x\)
Now, replace \(y\) back with \(\ln x\):
\[
\ln x = 3 \quad \text{or} \quad \ln x = -5
\]

### Step 4: Solve for \(x\)
To find \(x\), exponentiate both sides:
\[
x = e^3 \quad \text{or} \quad x = e^{-5}
\]

### Final Answer
The solutions are:
\[
x = e^3 \quad \text{or} \quad x = e^{-5}
\]

Would you like more details or have any questions? Here are some related problems to consider:

1. Solve \( (\ln x)^2 - 4\ln x + 3 = 0 \).
2. Solve \( e^{2\ln x} + 3e^{\ln x} - 4 = 0 \).
3. Solve \( \ln(x^2 + x) - \ln(x + 1) = 1 \).
4. Solve \( \ln(\ln x) + 2\ln x - 3 = 0 \).
5. Solve \( \ln(x) + \ln(2x-1) = 0 \).

**Tip:** Quadratic equations involving logarithms can often be simplified by substitution. This technique reduces the problem to solving a standard quadratic equation.

Discover the step-by-step solution to the logarithmic quadratic equation (lnx)^2 + 2lnx - 15 = 0. This detailed explanation covers important algebraic concepts, including logarithms and the quadratic formula, making it ideal for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation