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The equation given is:

\[
3^x + \frac{2}{3^{-x}} = 81
\]

We can start by simplifying the second term, \(\frac{2}{3^{-x}}\). This is equivalent to:

\[
\frac{2}{3^{-x}} = 2 \cdot 3^x
\]

Thus, the equation becomes:

\[
3^x + 2 \cdot 3^x = 81
\]

Now, combine like terms:

\[
3^x (1 + 2) = 81
\]

This simplifies to:

\[
3 \cdot 3^x = 81
\]

Dividing both sides by 3:

\[
3^x = \frac{81}{3} = 27
\]

Since \( 27 = 3^3 \), we can write:

\[
3^x = 3^3
\]

Thus, \(x = 3\).

The solution is \( x = 3 \).

Would you like a detailed breakdown of any step, or have any other questions?

Here are 5 related questions:
1. How do you simplify exponential expressions like \( \frac{1}{3^{-x}} \)?
2. What are the rules for solving equations involving powers with the same base?
3. How would the solution change if the base were different, e.g., 2 instead of 3?
4. How do you graph the function \( f(x) = 3^x + \frac{2}{3^{-x}} \)?
5. How would this equation change if \( 81 \) were replaced with another number?

**Tip:** Whenever you see an expression like \( \frac{1}{a^{-x}} \), remember that it can be rewritten as \( a^x \), which helps simplify calculations!

Solve the equation: 3^x + 2 / 3^(-x) = 81

Learn how to solve the exponential equation 3^x + 2 / 3^(-x) = 81 with step-by-step breakdowns of each step. This problem covers key concepts such as exponent rules and solving exponential equations. Suitable for students in Grades 9-11.

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