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The given problem asks for the number of real values of \(x\) that satisfy the equation:

\[
\frac{8^x + 27^x}{12^x + 18^x} = \frac{7}{6}
\]

Let's go step-by-step to solve this.

---

### Step 1: Express the terms in powers of simpler bases
Notice that:
- \(8 = 2^3\) and \(27 = 3^3\),
- \(12 = 2^2 \cdot 3\) and \(18 = 2 \cdot 3^2\).

So, rewrite the terms:
\[
8^x = (2^3)^x = 2^{3x}, \quad 27^x = (3^3)^x = 3^{3x}
\]
Similarly:
\[
12^x = (2^2 \cdot 3)^x = 2^{2x} \cdot 3^x, \quad 18^x = (2 \cdot 3^2)^x = 2^x \cdot 3^{2x}
\]

---

### Step 2: Simplify the equation
Now the equation becomes:

\[
\frac{2^{3x} + 3^{3x}}{2^{2x} \cdot 3^x + 2^x \cdot 3^{2x}} = \frac{7}{6}
\]

---

### Step 3: Substitution
Let \(a = 2^x\) and \(b = 3^x\). Then:

- \(2^{3x} = a^3\) and \(3^{3x} = b^3\),
- \(2^{2x} \cdot 3^x = a^2 \cdot b\) and \(2^x \cdot 3^{2x} = a \cdot b^2\).

Thus, the equation becomes:

\[
\frac{a^3 + b^3}{a^2 b + a b^2} = \frac{7}{6}
\]

---

### Step 4: Apply factorization
We know that:

\[
a^3 + b^3 = (a + b)(a^2 - ab + b^2)
\]

and

\[
a^2 b + a b^2 = ab(a + b).
\]

Thus, the equation becomes:

\[
\frac{(a + b)(a^2 - ab + b^2)}{ab(a + b)} = \frac{7}{6}.
\]

Since \(a + b \neq 0\) (as both \(a = 2^x\) and \(b = 3^x\) are positive), we can cancel \(a + b\) from the numerator and denominator:

\[
\frac{a^2 - ab + b^2}{ab} = \frac{7}{6}.
\]

---

### Step 5: Rewrite the equation
The equation now becomes:

\[
\frac{a^2}{ab} - \frac{ab}{ab} + \frac{b^2}{ab} = \frac{7}{6}.
\]

Simplifying each term:

\[
\frac{a}{b} + \frac{b}{a} - 1 = \frac{7}{6}.
\]

Let \(t = \frac{a}{b} = \frac{2^x}{3^x} = \left(\frac{2}{3}\right)^x\). Thus, the equation becomes:

\[
t + \frac{1}{t} - 1 = \frac{7}{6}.
\]

---

### Step 6: Solve for \(t\)
Rearrange the equation:

\[
t + \frac{1}{t} = \frac{13}{6}.
\]

Multiply both sides by \(t\):

\[
t^2 + 1 = \frac{13}{6} t.
\]

Rearrange into a standard quadratic form:

\[
6t^2 - 13t + 6 = 0.
\]

---

### Step 7: Solve the quadratic equation
Use the quadratic formula:

\[
t = \frac{-(-13) \pm \sqrt{(-13)^2 - 4 \cdot 6 \cdot 6}}{2 \cdot 6}.
\]

\[
t = \frac{13 \pm \sqrt{169 - 144}}{12} = \frac{13 \pm \sqrt{25}}{12}.
\]

\[
t = \frac{13 \pm 5}{12}.
\]

So:

\[
t = \frac{18}{12} = \frac{3}{2}, \quad t = \frac{8}{12} = \frac{2}{3}.
\]

---

### Step 8: Find corresponding values of \(x\)
Recall that \(t = \left(\frac{2}{3}\right)^x\).

- For \(t = \frac{3}{2}\):
  \[
  \left(\frac{2}{3}\right)^x = \frac{3}{2}.
  \]
  Taking logarithm on both sides:
  \[
  x \ln\left(\frac{2}{3}\right) = \ln\left(\frac{3}{2}\right).
  \]
  \[
  x = \frac{\ln\left(\frac{3}{2}\right)}{\ln\left(\frac{2}{3}\right)} = -1.
  \]

- For \(t = \frac{2}{3}\):
  \[
  \left(\frac{2}{3}\right)^x = \frac{2}{3}.
  \]
  This gives \(x = 1\).

---

### Step 9: Conclusion
The two solutions are \(x = -1\) and \(x = 1\). Thus, the number of real values of \(x\) is:

\[
\boxed{2}.
\]

---

Do you want further clarifications or additional details?  

Here are 5 related questions you could explore:
1. How do you solve other equations involving exponential terms like this?
2. What are some techniques to simplify equations using substitutions?
3. How do logarithms help in solving exponential equations?
4. What is the significance of the quadratic formula in problem-solving?
5. Can we solve such equations graphically?

**Tip:** When dealing with exponential equations, substitutions often help reduce the complexity.

Find the number of real values of x for which (8^x + 27^x) / (12^x + 18^x) = 7/6.

This page explains how to solve an exponential equation involving the terms (8^x + 27^x) / (12^x + 18^x) = 7/6. It includes key algebraic and logarithmic techniques, substitution steps, and quadratic formula application. The problem is suitable for students in Grades 10-12.

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