1.5 $$ 2 \sqrt{2-7 x}=\sqrt{-36 x} $$ $$ \sqrt{2-x}-x=-2 $$ .7 $$ 26-5^{2 x}=\left(5^{x}-6\right)^{2} $$ Show that the equation $$ 2^{2 x+1}+7.2^{x}-4=0 $$ has only ONE solution. SIMPLIFICATION

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To show that the equation

$$
2^{2x+1} + 7 \cdot 2^{x} - 4 = 0
$$

has only **one solution**, let's proceed with a step-by-step simplification and analysis.

### Step 1: Simplify the Equation

First, rewrite $$2^{2x + 1}$$ to make the equation easier to handle:

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

So, the original equation becomes:

$$
2 \cdot (2^{x})^2 + 7 \cdot 2^{x} - 4 = 0
$$

### Step 2: Substitute to Form a Quadratic Equation

Let’s make a substitution to simplify the equation further. Let:

$$
t = 2^{x}
$$

Since $$2^{x} > 0$$ for all real $$x$$, $$t > 0$$.

Substituting $$t$$ into the equation:

$$
2t^2 + 7t - 4 = 0
$$

### Step 3: Solve the Quadratic Equation

Now, solve for $$t$$ using the quadratic formula:

$$
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

Here, $$a = 2$$, $$b = 7$$, and $$c = -4$$. Plugging these values in:

$$
t = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm \sqrt{81}}{4} = \frac{-7 \pm 9}{4}
$$

This gives two potential solutions for $$t$$:

1. $$t = \frac{-7 + 9}{4} = \frac{2}{4} = \frac{1}{2}$$
2. $$t = \frac{-7 - 9}{4} = \frac{-16}{4} = -4$$

### Step 4: Consider the Valid Solutions

Recall that $$t = 2^{x} > 0$$. Therefore, $$t = -4$$ is **not valid**. Only $$t = \frac{1}{2}$$ is acceptable.

### Step 5: Find the Corresponding $$x$$

Solve for $$x$$ using the valid $$t$$:

$$
2^{x} = \frac{1}{2} \implies x = \log_{2}\left(\frac{1}{2}\right) = -1
$$

### Conclusion

The equation

$$
2^{2x+1} + 7 \cdot 2^{x} - 4 = 0
$$

has only **one valid solution**, which is:

$$
x = -1
$$

This single solution arises because the quadratic equation derived from the substitution yields only one positive value for $$t$$, corresponding to a unique real value for $$x$$.
The equation $$2^{2x+1} + 7 \cdot 2^{x} - 4 = 0$$ has only one solution, which is $$x = -1$$.

1.5

.7

Show that the equation has only ONE solution.
SIMPLIFICATION

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1.5 .7 Show that the equation has only ONE solution. SIMPLIFICATION