The problem asks us to compare two functions:

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

We are tasked with finding the row in the table where $f(x) = g(x)$.

Let's break down each function:

1. For $f(x) = 2(2)^x$:

   • This simplifies to $f(x) = 2^{x+1}$.

2. For $g(x) = \frac{1}{4}(4)^x$:

   • Since $4 = 2^2$, this becomes $g(x) = \frac{1}{4}(2^2)^x = \frac{1}{4}(2^{2x}) = 2^{2x-2}$.

We need to determine where $2^{x+1} = 2^{2x-2}$. Equating the exponents:

$x+1 = 2x-2$

Solving for $x$:

$x + 1 = 2x - 2$ $1 + 2 = 2x - x$ $x = 3$

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

Now, checking the tables provided:

• Left table (with $x$, $f(x)$, and $g(x)$):
  • When $x = 3$, $f(x) = 16$ and $g(x) = 16$.
  • This matches the condition $f(x) = g(x)$.

Therefore, the correct row is the one where $x = 3$, $f(x) = 16$, and $g(x) = 16$.

Do you want further clarification on this? Here are five questions to deepen your understanding:

1. How would the functions change if the base of $f(x)$ were different?
2. Can you generalize how to solve for $f(x) = g(x)$ with different exponential forms?
3. What are other methods to compare functions graphically or numerically?
4. How does the equality of exponents work when equating powers of the same base?
5. What if $g(x)$ had been a different base, like $3^x$? How would the solution approach change?

Tip: Always ensure you simplify functions with common bases when dealing with exponential equations—this can make solving much easier!