The equation provided is:

$4^x + 2^{x+2} = 3.$

Step-by-step solution:

1. Rewrite $4^x$: Recall that $4^x = (2^2)^x = 2^{2x}$. Thus, the equation becomes: $2^{2x} + 2^{x+2} = 3.$

2. Factor out $2^x$: Let $y = 2^x$ (where $y > 0$), so the equation becomes: $y^2 + 4y = 3.$

3. Simplify into a standard quadratic equation: Rearrange terms: $y^2 + 4y - 3 = 0.$

4. Solve the quadratic equation: Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = 4$, and $c = -3$. Substituting these values: $y = \frac{-4 \pm \sqrt{4^2 - 4(1)(-3)}}{2(1)} = \frac{-4 \pm \sqrt{16 + 12}}{2} = \frac{-4 \pm \sqrt{28}}{2}.$

5. Simplify further: $y = \frac{-4 \pm 2\sqrt{7}}{2} = -2 \pm \sqrt{7}.$

6. Choose the valid solution: Since $y = 2^x > 0$, we reject $y = -2 - \sqrt{7}$ (negative value). Thus: $y = -2 + \sqrt{7}.$

7. Find $x$: Recall $y = 2^x$. Thus: $2^x = -2 + \sqrt{7}.$ Taking the logarithm (base 2): $x = \log_2(-2 + \sqrt{7}).$

Final Answer:

$x = \log_2(-2 + \sqrt{7}).$

Would you like me to explain any specific step or verify with approximations?

Related Questions:

1. What is the approximate numerical value of $x$ using logarithms?
2. How do you solve equations with mixed exponential terms like $4^x + 2^{x+2} = 3$?
3. What does the quadratic substitution technique simplify in exponential equations?
4. Can $y = -2 + \sqrt{7}$ be approximated for practical computation?
5. Why must $2^x > 0$ for all real values of $x$?

Tip:

Always check the domain of logarithmic and exponential equations to ensure that solutions make sense (e.g., no negative bases for logs).