Solving Exponential Equations Find The Truth Set Of 2^(2x+2) - 5(2^x) + 1 = 0

Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of exponential equations. We're going to tackle the equation $2^{2x+2} - 5(2^x) + 1 = 0$ and uncover its truth set within the realm of real numbers. This equation might look a bit intimidating at first glance, but don't worry, we'll break it down step-by-step and make it super easy to understand. So, grab your thinking caps, and let's get started!

1. Laying the Foundation: Understanding Exponential Equations

Before we jump into solving our specific equation, let's take a moment to appreciate the beauty and power of exponential equations. Exponential equations are those where the variable appears in the exponent. They pop up in various real-world scenarios, from modeling population growth to calculating compound interest. The key to solving these equations lies in understanding the properties of exponents and using clever algebraic manipulations.

When you're dealing with exponential equations, remember that the goal is usually to isolate the exponential term. This often involves using techniques like substitution or logarithms. In our case, we'll employ a substitution method to transform the equation into a more manageable form.

2. Transforming the Equation: A Clever Substitution

Alright, let's get our hands dirty with the equation at hand: $2^{2x+2} - 5(2^x) + 1 = 0$. The first thing we notice is the presence of $2^x$ in multiple terms. This hints at a possible simplification using substitution. Let's make a substitution: let $y = 2^x$. This seemingly simple move will make a world of difference.

Now, let's rewrite the equation in terms of $y$. First, we can rewrite $2^{2x+2}$ as $2^{2x} imes 2^2$, which is the same as $4 imes (2^x)^2$. Since $y = 2^x$, we have $(2^x)^2 = y^2$. Thus, $2^{2x+2}$ becomes $4y^2$.

Substituting $y = 2^x$ and $4y^2$ for $2^{2x+2}$ into the original equation, we get a quadratic equation in terms of $y$: $4y^2 - 5y + 1 = 0$. Wow, that looks much friendlier, doesn't it? This transformation is a crucial step in solving the equation. By changing the variable, we've turned a complex exponential equation into a familiar quadratic equation.

3. Solving the Quadratic Equation: Unveiling the Roots

Now that we have the quadratic equation $4y^2 - 5y + 1 = 0$, we can employ our trusty quadratic equation-solving skills. There are several ways to tackle this: factoring, completing the square, or using the quadratic formula. In this case, factoring seems like a viable option. Factoring is a great method because it allows you to break down the quadratic into simpler terms, making it easier to find the solutions.

We need to find two numbers that multiply to give $4 imes 1 = 4$ and add up to $-5$. Those numbers are $-4$ and $-1$. So, we can rewrite the middle term as $-5y = -4y - y$. This gives us:

$4y^2 - 4y - y + 1 = 0$

Now, we factor by grouping:

$4y(y - 1) - 1(y - 1) = 0$

$(4y - 1)(y - 1) = 0$

Setting each factor equal to zero, we get two possible solutions for $y$:

4y - 1 = 0 => y = rac{1}{4}

$y - 1 = 0 => y = 1$

So, the solutions for $y$ are rac{1}{4} and $1$. These are the roots of our quadratic equation, but remember, we're not done yet. We need to find the values of $x$ that correspond to these values of $y$.

4. Back to the Original Variable: Finding the Truth Set for $x$

We've found the solutions for $y$, but our ultimate goal is to find the values of $x$ that satisfy the original equation. Remember our substitution: $y = 2^x$. Now we need to substitute back and solve for $x$.

For y = rac{1}{4}, we have 2^x = rac{1}{4}. We can rewrite rac{1}{4} as $2^{-2}$. So, we have:

$2^x = 2^{-2}$

Since the bases are equal, the exponents must be equal. Therefore, $x = -2$.

For $y = 1$, we have $2^x = 1$. We can rewrite $1$ as $2^0$. So, we have:

$2^x = 2^0$

Again, since the bases are equal, the exponents must be equal. Therefore, $x = 0$.

We've found two solutions for $x$: $x = -2$ and $x = 0$. These are the values of $x$ that make the original equation true. These solutions form the truth set of the equation.

5. The Truth Set: Our Final Answer

We've navigated through the exponential equation, made a clever substitution, solved a quadratic equation, and finally arrived at the truth set for $x$. The truth set is the set of all values of $x$ that satisfy the original equation. In our case, the truth set is $\{-2, 0\}$.

So, the truth set of the equation $2^{2x+2} - 5(2^x) + 1 = 0$, where $x$ belongs to the set of real numbers, is $\{-2, 0\}$.

Conclusion: A Triumph Over Exponential Equations

There you have it, guys! We've successfully unraveled the truth set of the exponential equation $2^{2x+2} - 5(2^x) + 1 = 0$. We started by understanding the basics of exponential equations, then we used a substitution to transform the equation into a more manageable quadratic form. We solved the quadratic equation, substituted back to find the values of $x$, and finally determined the truth set.

Remember, the key to solving these types of problems is to break them down into smaller, more manageable steps. Don't be intimidated by the complexity of the equation; instead, look for ways to simplify it using algebraic techniques. With practice and a solid understanding of the underlying concepts, you can conquer any exponential equation that comes your way. Keep exploring, keep learning, and keep having fun with math!

Find the solution set of the equation $2^{2x+2} - 5(2^x) + 1 = 0$ for $x$ belonging to the set of real numbers.

Solving Exponential Equations Find the Truth Set of 2^(2x+2) - 5(2^x) + 1 = 0