Solve the following exponential equation. Express irrational solutions in exact form. $$ 2^{2 x}+2^{x+1}-35=0 $$

Question

UpStudy AI · Accepted Answer

To solve the exponential equation:

$$
2^{2x} + 2^{x+1} - 35 = 0
$$

**Step 1: Substitution**

Let $$ y = 2^x $$. Then:

$$
2^{2x} = (2^x)^2 = y^2 \
2^{x+1} = 2 \cdot 2^x = 2y
$$

Substituting these into the original equation gives:

$$
y^2 + 2y - 35 = 0
$$

**Step 2: Solve the Quadratic Equation**

Solve for $$ y $$:

$$
y^2 + 2y - 35 = 0
$$

Using the quadratic formula:

$$
y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-35)}}{2} = \frac{-2 \pm \sqrt{4 + 140}}{2} = \frac{-2 \pm \sqrt{144}}{2} = \frac{-2 \pm 12}{2}
$$

This gives two potential solutions:

$$
y = \frac{10}{2} = 5 \quad \text{or} \quad y = \frac{-14}{2} = -7
$$

Since $$ y = 2^x $$ and $$ 2^x $$ is always positive, $$ y = -7 $$ is not a valid solution. Therefore, we have:

$$
2^x = 5
$$

**Step 3: Solve for $$ x $$**

Taking the logarithm base 2 of both sides:

$$
x = \log_2 5
$$

This is the exact form of the solution.

**Answer:**  
After simplifying, the solution is log base 2 of five. In symbols,

x = log₂ 5
The solution is $$ x = \log_2 5 $$.

Solve the following exponential equation. Express irrational solutions in exact form.