The problem asks us to solve the exponential equation $4 \times 2^{2x+1} = 80$ for $x$.

1. Problem Description

The problem asks us to solve the exponential equation $4 \times 2^{2x+1} = 80$ for $x$.

2. Solution Steps

First, divide both sides of the equation by 4:

$4 \times 2^{2x+1} = 80$

$2^{2x+1} = \frac{80}{4}$

$2^{2x+1} = 20$

Next, we can take the logarithm of both sides. Let's use the base-2 logarithm.

$log_2(2^{2x+1}) = log_2(20)$

Using the logarithm property $log_b(a^c) = c \cdot log_b(a)$, we get:

$(2x+1)log_2(2) = log_2(20)$

Since $log_2(2) = 1$, we have:

$2x+1 = log_2(20)$

Subtract 1 from both sides:

$2x = log_2(20) - 1$

Divide both sides by 2:

$x = \frac{log_2(20) - 1}{2}$

We can express $log_2(20)$ as $log_2(4 \times 5) = log_2(4) + log_2(5) = log_2(2^2) + log_2(5) = 2 + log_2(5)$.

Therefore,

$x = \frac{2 + log_2(5) - 1}{2} = \frac{1 + log_2(5)}{2}$

We can rewrite this as:

$x = \frac{1}{2} + \frac{log_2(5)}{2} = \frac{1}{2} + \frac{1}{2}log_2(5) = \frac{1}{2} + log_2(5^{1/2}) = \frac{1}{2} + log_2(\sqrt{5})$.

Alternatively, we can find an approximate value using a calculator.

$x = \frac{log_2(20) - 1}{2}$. We know that $2^4=16$ and $2^5=32$, so $log_2(20)$ is between 4 and

5. Using a calculator, $log_2(20) \approx 4.3219$.

$x \approx \frac{4.3219 - 1}{2} = \frac{3.3219}{2} \approx 1.66095$

3. Final Answer

$x = \frac{log_2(20) - 1}{2} = \frac{1 + log_2(5)}{2} = \frac{1}{2} + log_2(\sqrt{5})$