https://file.aiquickdraw.com/mathbot/file/1727363989586cy6msnxb.jpg

The equation in the image is:

\[
2^{2 - x} - 5 \cdot 2^x - 5 = 0
\]

Let's solve it step-by-step.

### Step 1: Simplify the equation
We begin by rewriting \( 2^{2 - x} \) using the property of exponents:
\[
2^{2 - x} = 2^2 \cdot 2^{-x} = 4 \cdot \frac{1}{2^x}.
\]
Substituting this back into the equation:

\[
4 \cdot \frac{1}{2^x} - 5 \cdot 2^x - 5 = 0.
\]

### Step 2: Substitution
Let \( y = 2^x \). Then the equation becomes:

\[
4 \cdot \frac{1}{y} - 5y - 5 = 0.
\]

### Step 3: Eliminate the fraction
Multiply the entire equation by \( y \) to eliminate the fraction:

\[
4 - 5y^2 - 5y = 0.
\]

### Step 4: Rearrange into standard form
This is a quadratic equation. Rearrange it into standard form:

\[
-5y^2 - 5y + 4 = 0.
\]

### Step 5: Solve the quadratic equation
Multiply through by \( -1 \) to make the leading coefficient positive:

\[
5y^2 + 5y - 4 = 0.
\]

Now, use the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
where \( a = 5 \), \( b = 5 \), and \( c = -4 \).

Substitute these values into the quadratic formula:

\[
y = \frac{-5 \pm \sqrt{5^2 - 4(5)(-4)}}{2(5)} = \frac{-5 \pm \sqrt{25 + 80}}{10} = \frac{-5 \pm \sqrt{105}}{10}.
\]

So:
\[
y = \frac{-5 \pm \sqrt{105}}{10}.
\]

### Step 6: Approximate the solutions
Now, approximate \( \sqrt{105} \):

\[
\sqrt{105} \approx 10.247.
\]

Thus, the two possible values for \( y \) are:

\[
y_1 = \frac{-5 + 10.247}{10} \approx 0.5247,
\]
\[
y_2 = \frac{-5 - 10.247}{10} \approx -1.5247.
\]

### Step 7: Solve for \( x \)
Recall that \( y = 2^x \). So, for \( y_1 = 0.5247 \):

\[
2^x = 0.5247 \quad \Rightarrow \quad x = \log_2(0.5247).
\]

For \( y_2 = -1.5247 \), since \( 2^x \) is always positive, there is no solution for this case.

### Step 8: Calculate the logarithm
We calculate \( x \) for \( 2^x = 0.5247 \):

\[
x = \log_2(0.5247) = \frac{\log(0.5247)}{\log(2)}.
\]

Let me compute the final value.The solution for \( x \) is approximately \( x \approx -0.93 \).

### Final Answer:
The solution to the equation \( 2^{2 - x} - 5 \cdot 2^x - 5 = 0 \) is:

\[
x \approx -0.93.
\]

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How does the quadratic formula work in solving quadratic equations?
2. Why do we need to reject the negative solution for \( y \) in this context?
3. Can logarithms of negative numbers be defined in real numbers?
4. How does the substitution method simplify solving exponential equations?
5. What are some other methods to solve exponential equations?

**Tip:** Always check if the final values make sense in the context of the original problem. For instance, exponential expressions cannot result in negative values.

Solve the equation 2^(2-x) - 5*2^x - 5 = 0.

This page provides a step-by-step solution to the equation 2^(2-x) - 5*2^x - 5 = 0. Concepts covered include exponential equations, the substitution method, the quadratic formula, and logarithms. Suitable for students in Grades 10-12.

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