The functions given in the image are:

$f(x) = 2(2)^x$ $g(x) = \frac{1}{4}(4)^x$

We are asked to find the row in the table where $f(x) = g(x)$.

Let's solve this analytically first.

We need to find when:

$2(2)^x = \frac{1}{4}(4)^x$

First, simplify the equation:

$2(2)^x = \frac{1}{4}(2^2)^x = \frac{1}{4}(2^{2x})$

Multiply both sides by 4:

$8(2)^x = (2)^{2x}$

Now express both sides with the same base:

$2^3(2)^x = 2^{2x}$

Simplify the left-hand side:

$2^{x+3} = 2^{2x}$

Since the bases are the same, equate the exponents:

$x + 3 = 2x$

Solve for $x$:

$3 = x$

So, $f(x) = g(x)$ when $x = 3$.

Now, let's check the values from the table.

For $x = 3$:

• $f(3) = 2(2)^3 = 2 \times 8 = 16$
• $g(3) = \frac{1}{4}(4)^3 = \frac{1}{4} \times 64 = 16$

Both $f(3) = g(3) = 16$, so the correct row is $x = 3$.

Based on the two tables in the image, the row that corresponds to $x = 3$ is in the second table.

Would you like more details or have any questions? Here are some related questions to expand your understanding:

1. How would you solve this if the functions involved different bases?
2. What happens if the base of one of the functions is negative?
3. How can you graph $f(x)$ and $g(x)$ to visually confirm the solution?
4. What are some general methods to solve exponential equations algebraically?
5. How does this method change if $f(x)$ and $g(x)$ were logarithmic functions?

Tip: When solving equations with exponential functions, always try to express both sides with the same base for easier comparison.