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solve it without taking out a common term

To solve the equation \( 9^{x+1} + 4^{x+1} - 6^{x+1} = 7 \cdot 6^x \) without taking out a common term, we can approach it step by step. 

### Step 1: Express the terms in a simpler form
First, let's express \( 9^{x+1} \), \( 4^{x+1} \), and \( 6^{x+1} \) as products of powers:
\[
9^{x+1} = 9 \cdot 9^x, \quad 4^{x+1} = 4 \cdot 4^x, \quad \text{and} \quad 6^{x+1} = 6 \cdot 6^x.
\]
So the equation becomes:
\[
9 \cdot 9^x + 4 \cdot 4^x - 6 \cdot 6^x = 7 \cdot 6^x.
\]

### Step 2: Isolate the exponential terms
Now, let's rewrite the equation:
\[
9 \cdot 9^x + 4 \cdot 4^x - 6 \cdot 6^x - 7 \cdot 6^x = 0.
\]
Simplifying the terms involving \( 6^x \) gives us:
\[
9 \cdot 9^x + 4 \cdot 4^x - 13 \cdot 6^x = 0.
\]

### Step 3: Solve the equation
At this point, the equation is not easily factorable, and solving it directly would typically require either numerical methods or some specific insight into the values of \( x \). However, a possible substitution might help:

#### Substitute \( x = 0 \):
\[
9 \cdot 9^0 + 4 \cdot 4^0 - 13 \cdot 6^0 = 9 \cdot 1 + 4 \cdot 1 - 13 \cdot 1 = 9 + 4 - 13 = 0.
\]
So \( x = 0 \) is a solution.

### Final Answer:
The solution is \( x = 0 \).

Would you like further details or have any questions? Here are some related problems for practice:

1. Solve the equation \( 2^{x+1} + 3^{x+1} - 5^{x+1} = 8 \cdot 5^x \) without taking out a common term.
2. How would you solve \( 10^{x+1} + 2^{x+1} - 3^{x+1} = 4 \cdot 3^x \)?
3. What happens if you substitute \( x = 1 \) into the original equation?
4. Can the original equation have multiple solutions? Explain.
5. How does the equation change if we alter the constant on the right-hand side?

**Tip:** When solving exponential equations, substituting small integer values of \( x \) (like 0, 1, or -1) can often reveal simple solutions quickly.

Learn how to solve the exponential equation 9^(x+1) + 4^(x+1) - 6^(x+1) = 7 * 6^x without factoring out a common term. This problem explores properties of exponents and requires understanding of exponential rules and identities. Suitable for high school students.

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