b.) $$ 2^{2 x+1}+4^{x+1}=2^{2 x}\left(\frac{9}{2}\right) $$

Question

UpStudy AI · Accepted Answer

To solve the equation:

$$ 2^{2x + 1} + 4^{x + 1} = 2^{2x} \left(\frac{9}{2}\right) $$

Let's follow these steps:

1. **Express all terms with the same base:**

Since $$4 = 2^2$$, we can rewrite $$4^{x + 1}$$ as:
   $$
   4^{x + 1} = (2^2)^{x + 1} = 2^{2(x + 1)} = 2^{2x + 2}
   $$
   
   So the equation becomes:
   $$
   2^{2x + 1} + 2^{2x + 2} = \frac{9}{2} \cdot 2^{2x}
   $$

2. **Factor out the common term $$2^{2x}$$:**
   $$
   2^{2x} \cdot 2^1 + 2^{2x} \cdot 2^2 = \frac{9}{2} \cdot 2^{2x}
   $$
   $$
   2^{2x} (2 + 4) = \frac{9}{2} \cdot 2^{2x}
   $$
   $$
   2^{2x} \cdot 6 = \frac{9}{2} \cdot 2^{2x}
   $$

3. **Simplify the equation:**
   
   Since $$2^{2x} \neq 0$$ for any real $$x$$, we can divide both sides by $$2^{2x}$$:
   $$
   6 = \frac{9}{2}
   $$
   
4. **Analyze the result:**
   
   The equation simplifies to $$6 = 4.5$$, which is a **contradiction**. This means there's no real value of $$x$$ that satisfies the original equation.

**Conclusion:**  
There is **no real solution** to the given equation.

**Final Answer:**  
No real solution exists for the equation; there is no value of x that satisfies it.
There is no real solution to the equation $$2^{2x+1} + 4^{x+1} = 2^{2x} \left(\frac{9}{2}\right)$$.

b.) \( 2^{2 x+1}+4^{x+1}=2^{2 x}\left(\frac{9}{2}\right) \)

Real Tutor Solution

Answer

Solution

Extra Insights

Related Questions

Latest Algebra Questions