To solve the inequality:

\[
\log_{(5x+3)}(2x^2) \leq 1
\]

Let's go through the steps systematically:

### Step 1: Understand the inequality

The inequality involves a logarithm with base \(5x + 3\) and argument \(2x^2\). For the logarithm to be defined:
1. The base \(5x + 3 > 0\) and \(5x + 3 \neq 1\).
2. The argument \(2x^2 > 0\).

### Step 2: Set the domain

1. **Base condition**: 
   \[
   5x + 3 > 0 \implies x > -\frac{3}{5}
   \]
   And, \(5x + 3 \neq 1\), so:
   \[
   5x + 3 \neq 1 \implies x \neq -\frac{2}{5}.
   \]

2. **Argument condition**: 
   \[
   2x^2 > 0 \implies x \neq 0.
   \]

Combining these, the domain is:

\[
x > -\frac{3}{5}, \, x \neq -\frac{2}{5}, \, x \neq 0.
\]

### Step 3: Transform the inequality

Rewrite the inequality:

\[
\log_{(5x+3)}(2x^2) \leq 1
\]

Using the property of logarithms, \( \log_b(a) \leq c \implies a \leq b^c \) when \(b > 1\), and \( a \geq b^c \) when \(0 < b < 1\).

Let's analyze two cases based on the base \(5x + 3\):

#### Case 1: \(5x + 3 > 1\)

This implies \(x > -\frac{2}{5}\).

For \(5x + 3 > 1\), the logarithm inequality becomes:

\[
2x^2 \leq (5x + 3)^1.
\]

Simplifying:

\[
2x^2 \leq 5x + 3.
\]

Rearrange to form a quadratic inequality:

\[
2x^2 - 5x - 3 \leq 0.
\]

#### Case 2: \(0 < 5x + 3 < 1\)

This implies \(-\frac{3}{5} < x < -\frac{2}{5}\).

For \(0 < 5x + 3 < 1\), the logarithm inequality becomes:

\[
2x^2 \geq 5x + 3.
\]

Rearrange to form another quadratic inequality:

\[
2x^2 - 5x - 3 \geq 0.
\]

### Step 4: Solve the quadratic inequalities

Let's solve each quadratic inequality separately.

1. **Quadratic for Case 1**: 
   \[
   2x^2 - 5x - 3 = 0.
   \]

Find the roots using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4}.
\]

\[
x = \frac{5 \pm 7}{4}.
\]

So, the roots are:

\[
x_1 = \frac{12}{4} = 3, \quad x_2 = \frac{-2}{4} = -\frac{1}{2}.
\]

The quadratic \(2x^2 - 5x - 3 \leq 0\) will be negative or zero between the roots:

\[
-\frac{1}{2} \leq x \leq 3.
\]

But we must also consider the domain restriction \(x > -\frac{2}{5}\), so:

\[
-\frac{2}{5} < x \leq 3.
\]

2. **Quadratic for Case 2**:
   \[
   2x^2 - 5x - 3 \geq 0.
   \]

Using the same roots \(x_1 = -\frac{1}{2}\) and \(x_2 = 3\), the quadratic will be positive outside the interval between the roots:

\[
x \leq -\frac{1}{2} \quad \text{or} \quad x \geq 3.
\]

However, since \( -\frac{3}{5} < x < -\frac{2}{5} \), we intersect this with the domain:

\[
-\frac{3}{5} < x < -\frac{2}{5}.
\]

### Step 5: Combine the intervals

The solution intervals from both cases are:

\[
\left( -\frac{2}{5}, 3 \right] \cup \left( -\frac{3}{5}, -\frac{2}{5} \right).
\]

### Final Answer

The solution to the inequality is:

\[
-\frac{3}{5} < x \leq 3, \, x \neq -\frac{2}{5}.
\]

Would you like further clarification on any of the steps? Here are some related questions:

1. What is the method to solve quadratic inequalities?
2. How do logarithmic inequalities change based on the base value?
3. Why do we exclude certain values from the domain?
4. How does the quadratic formula work?
5. What are the properties of logarithms that are useful in inequalities?

**Tip:** Always check the domain restrictions when solving logarithmic inequalities to avoid undefined expressions.

Learn how to solve the logarithmic inequality log(5x+3)(2x^2)<=1 with a detailed step-by-step solution. This problem involves logarithms and quadratic inequalities, covering key concepts such as domain restrictions and solving quadratic equations. Suitable for students in Grades 10-12.

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