To solve the equation $4^x = 2^x + 1$, follow these steps:

Step 1: Express $4^x$ in terms of $2^x$

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

Step 2: Let $y = 2^x$

Substitute $y$ for $2^x$, which gives us: $y^2 = y + 1$

Step 3: Solve the quadratic equation

The equation $y^2 = y + 1$ can be rewritten as: $y^2 - y - 1 = 0$ This is a quadratic equation in $y$, and we can solve it using the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For our equation $y^2 - y - 1 = 0$, the coefficients are $a = 1$, $b = -1$, and $c = -1$. Plugging these into the quadratic formula gives: $y = \frac{1 \pm \sqrt{1^2 + 4 \times 1 \times 1}}{2 \times 1} = \frac{1 \pm \sqrt{5}}{2}$

Step 4: Solve for $x$

We have $y = 2^x$, so: $2^x = \frac{1 + \sqrt{5}}{2} \quad \text{or} \quad 2^x = \frac{1 - \sqrt{5}}{2}$

However, $2^x$ must be positive, and since $\frac{1 - \sqrt{5}}{2}$ is negative, we discard that solution.

Thus, the solution is: $2^x = \frac{1 + \sqrt{5}}{2}$ Taking the logarithm on both sides: $x = \log_2 \left(\frac{1 + \sqrt{5}}{2}\right)$

Final Answer

$x = \log_2 \left(\frac{1 + \sqrt{5}}{2}\right)$ This is the exact solution for $x$.

Would you like further details or have any questions?

Here are 5 related questions for further exploration:

1. How do you solve quadratic equations using the quadratic formula?
2. What is the significance of the golden ratio in the solution $\frac{1 + \sqrt{5}}{2}$?
3. How do logarithmic functions work and how are they applied in solving equations?
4. Can you explore how to solve similar exponential equations where bases are different?
5. What are some examples of equations where the discriminant is negative, and how are those solved?

Tip: Always check the domain of the solutions, especially when dealing with logarithms or square roots.