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We are tasked with finding \(\left(\frac{3}{2}\right)^x\) given the equation \(4^x + 6^x = 9^x\).

### Solution:
We begin by analyzing the given equation:
\[
4^x + 6^x = 9^x.
\]

First, observe that \(4\), \(6\), and \(9\) are powers of smaller numbers. Specifically:
\[
4 = 2^2, \quad 6 = 2 \cdot 3, \quad 9 = 3^2.
\]
However, instead of directly expanding powers, let us test small integer values of \(x\) to see if any satisfy this equation.

#### Step 1: Test \(x = 1\):
\[
4^1 + 6^1 = 9^1,
\]
which gives:
\[
4 + 6 = 9 \quad \Rightarrow \quad 10 \neq 9.
\]
So, \(x = 1\) is not a solution.

#### Step 2: Test \(x = 2\):
\[
4^2 + 6^2 = 9^2,
\]
which gives:
\[
16 + 36 = 81 \quad \Rightarrow \quad 52 \neq 81.
\]
Thus, \(x = 2\) is also not a solution.

#### Step 3: Test \(x = 0\):
\[
4^0 + 6^0 = 9^0,
\]
which gives:
\[
1 + 1 = 1 \quad \Rightarrow \quad 2 \neq 1.
\]
So, \(x = 0\) is not a solution either.

#### Step 4: Test \(x = 3\):
\[
4^3 + 6^3 = 9^3,
\]
which gives:
\[
64 + 216 = 729 \quad \Rightarrow \quad 280 \neq 729.
\]
Thus, \(x = 3\) is not a solution.

#### Step 5: Test \(x = 0.5\):
We now try \(x = \frac{1}{2}\) (which is a non-integer value):
\[
4^{1/2} + 6^{1/2} = 9^{1/2},
\]
which simplifies to:
\[
2 + \sqrt{6} \neq 3.
\]
Thus, this is not a solution.

#### Step 6: Try a logarithmic method:
We attempt to take logarithms on both sides of the equation \(4^x + 6^x = 9^x\), but the current approach can be complicated.

We will next work to solve it more cleanly. Try

This problem involves solving the equation 4^x + 6^x = 9^x and finding the value of (3/2)^x. Key concepts include exponentiation and testing values. Suitable for high school students, Grades 10-12.

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