Find A And B: Square Root Equation

This article delves into a fascinating mathematical problem: finding the values of $a$ and $b$ that make the equation $\sqrt{648}=\sqrt{2^a \cdot 3^b}$ true. We'll break down this problem step-by-step, exploring the concepts of prime factorization and square roots to arrive at the correct solution. Understanding these fundamental mathematical principles is crucial for solving a wide range of algebraic and number theory problems. Whether you're a student tackling homework, a curious mind exploring mathematics, or preparing for standardized tests, this guide will provide clarity and a solid foundation.

Let's start by dissecting the equation $\sqrt{648}=\sqrt{2^a \cdot 3^b}$. Our primary goal is to determine the specific integer values for $a$ and $b$ that satisfy this equality. To do this effectively, we need to express the number 648 in its prime factorized form. Prime factorization is the process of breaking down a composite number into its prime number components – numbers greater than 1 that are only divisible by 1 and themselves (like 2, 3, 5, 7, 11, etc.). This is a fundamental technique in number theory, allowing us to understand the building blocks of any integer. Once we have the prime factorization of 648, we can compare it to the prime factorization on the right side of the equation, $2^a \cdot 3^b$. The square root on both sides also plays a critical role. Recall that the square root of a number raised to a power can be simplified. Specifically, $\sqrt{x^n} = x^{n/2}$. Applying this to our equation, we get $648^{1/2} = (2^a \cdot 3^b)^{1/2}$, which simplifies further to $648^{1/2} = 2^{a/2} \cdot 3^{b/2}$. Our task is to find $a$ and $b$ such that $648 = 2^a \cdot 3^b$.

To begin, let's find the prime factorization of 648. We can start by dividing 648 by the smallest prime number, 2:

• $648 \div 2 = 324$
• $324 \div 2 = 162$
• $162 \div 2 = 81$

Now, 81 is not divisible by 2. The next prime number is 3. Let's see if 81 is divisible by 3:

• $81 \div 3 = 27$
• $27 \div 3 = 9$
• $9 \div 3 = 3$
• $3 \div 3 = 1$

We have reached 1, so we have completed the prime factorization. Combining all the prime factors, we find that $648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3$. In exponential form, this is $648 = 2^3 \cdot 3^4$.

Now, let's return to our original equation: $\sqrt{648}=\sqrt{2^a \cdot 3^b}$. Since we've found that $648 = 2^3 \cdot 3^4$, we can substitute this into the equation:

$\sqrt{2^3 \cdot 3^4} = \sqrt{2^a \cdot 3^b}$

For this equation to be true, the prime factorizations inside the square roots must be identical. This means that the exponents of the corresponding prime bases must be equal. Therefore, by comparing the exponents:

• The exponent of 2 on the left side is 3, and on the right side, it is $a$. So, $a = 3$.
• The exponent of 3 on the left side is 4, and on the right side, it is $b$. So, $b = 4$.

Thus, the values that make the equation true are $a=3$ and $b=4$. This corresponds to option B in the given choices.

Understanding Prime Factorization and Square Roots

To solidify our understanding, let's delve a bit deeper into the concepts we've used. Prime factorization is a cornerstone of number theory. It allows us to represent any integer greater than 1 as a unique product of prime numbers. For example, the number 12 can be uniquely expressed as $2^2 \cdot 3^1$. This uniqueness, known as the Fundamental Theorem of Arithmetic, is incredibly powerful. It forms the basis for algorithms in cryptography, simplifies the process of finding the greatest common divisor (GCD) and least common multiple (LCM) of numbers, and is essential for solving equations like the one we tackled. When we factorize 648 into $2^3 \cdot 3^4$, we are essentially revealing its fundamental structure. This structured representation makes it easy to manipulate and compare with other numbers or expressions.

Square roots are another fundamental mathematical operation. The square root of a non-negative number $x$, denoted by $\sqrt{x}$, is a number $y$ such that $y^2 = x$. For example, $\sqrt{9} = 3$ because $3^2 = 9$. When dealing with prime factorizations under a square root, we can utilize the property $(x^m \cdot y^n)^{1/2} = x^{m/2} \cdot y^{n/2}$. Applying this to our problem, $\sqrt{2^3 \cdot 3^4} = 2^{3/2} \cdot 3^{4/2} = 2^{1.5} \cdot 3^2$. However, the equation is given as $\sqrt{648}=\sqrt{2^a \cdot 3^b}$, which implies that $2^a \cdot 3^b$ should be the number inside the square root, not the result of applying the square root. Therefore, we are comparing the radicands (the numbers inside the square root symbol). We found $648 = 2^3 \cdot 3^4$. The right side of the equation is $\sqrt{2^a \cdot 3^b}$. For these to be equal, the numbers inside the square roots must be equal: $648 = 2^a \cdot 3^b$. By direct comparison of the prime factorizations, we must have $a=3$ and $b=4$. This approach avoids potential confusion with fractional exponents arising from the square root operation itself, focusing directly on equating the terms within the radical.

Step-by-Step Solution

Let's reiterate the process clearly:

1. Identify the equation: We are given $\sqrt{648}=\sqrt{2^a \cdot 3^b}$.
2. Simplify the left side: The core of the problem lies in understanding the prime factorization of 648. We performed this by repeatedly dividing by prime numbers:
   • $648 = 2 \times 324$
   • $324 = 2 \times 162$
   • $162 = 2 \times 81$
   • $81 = 3 \times 27$
   • $27 = 3 \times 9$
   • $9 = 3 \times 3$ So, $648 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 = 2^3 \cdot 3^4$.
3. Equate the radicands: Since the square roots on both sides are equal, the numbers inside the square roots must also be equal. Therefore, $648 = 2^a \cdot 3^b$.
4. Substitute the prime factorization: Substitute the prime factorization of 648 into the equation: $2^3 \cdot 3^4 = 2^a \cdot 3^b$.
5. Compare exponents: For the equality to hold true, the exponents of the corresponding prime bases on both sides must be identical. This is a direct application of the uniqueness of prime factorization.
   • Comparing the exponents of 2: $3 = a$.
   • Comparing the exponents of 3: $4 = b$.
6. State the solution: The values of $a$ and $b$ that satisfy the equation are $a=3$ and $b=4$.

This detailed breakdown ensures that the logic is transparent and easy to follow. Each step builds upon the previous one, leading directly to the correct answer. This method is robust and can be applied to similar problems involving prime factorization and exponents.

Exploring Alternative Options and Common Pitfalls

When faced with multiple-choice questions like this, it's beneficial to understand why incorrect options are wrong. Let's consider the given options:

• A. $a=3, b=2$: If $a=3$ and $b=2$, then $2^a \cdot 3^b = 2^3 \cdot 3^2 = 8 \cdot 9 = 72$. Clearly, $\sqrt{72} \neq \sqrt{648}$.
• B. $a=3, b=4$: If $a=3$ and $b=4$, then $2^a \cdot 3^b = 2^3 \cdot 3^4 = 8 \cdot 81 = 648$. This matches our calculation, so $\sqrt{648} = \sqrt{2^3 \cdot 3^4}$ is true.
• C. $a=4, b=3$: If $a=4$ and $b=3$, then $2^a \cdot 3^b = 2^4 \cdot 3^3 = 16 \cdot 27 = 432$. Clearly, $\sqrt{432} \neq \sqrt{648}$.
• D. $a=2, b=3$: If $a=2$ and $b=3$, then $2^a \cdot 3^b = 2^2 \cdot 3^3 = 4 \cdot 27 = 108$. Clearly, $\sqrt{108} \neq \sqrt{648}$.

As you can see, only option B yields the correct value for 648 when plugged into the expression $2^a \cdot 3^b$.

A common pitfall in problems like this is misunderstanding the role of the square root. Some might incorrectly try to divide the exponents of the prime factorization of 648 by 2, assuming they are solving for the exponents outside the square root. For instance, if one incorrectly thought they needed to find $a$ and $b$ such that $\sqrt{648} = 2^a \cdot 3^b$, they might take $648 = 2^3 \cdot 3^4$ and try to get $a = 3/2$ and $b = 4/2 = 2$. However, the equation is $\sqrt{648}=\sqrt{2^a \cdot 3^b}$, meaning we are directly comparing the radicands. Therefore, the exponents must match directly.

Another potential error is in the prime factorization itself. A mistake in dividing or identifying prime numbers could lead to an incorrect prime factorization of 648, which would then propagate through the rest of the calculation. Always double-check your prime factorization steps to ensure accuracy. The process of breaking down 648 into its prime factors, $2^3 \cdot 3^4$, is the most critical initial step. Ensuring this is correct is paramount.

Conclusion

We have successfully navigated the problem of finding the values of $a$ and $b$ that satisfy the equation $\sqrt{648}=\sqrt{2^a \cdot 3^b}$. By employing the powerful technique of prime factorization, we determined that $648 = 2^3 \cdot 3^4$. Comparing this to the expression $2^a \cdot 3^b$ within the square root on the right side of the equation, we concluded that $a=3$ and $b=4$. This solution highlights the elegance and utility of fundamental mathematical concepts. Mastering prime factorization and understanding how exponents work within radicals are skills that will serve you well in various mathematical endeavors.

For further exploration into number theory and algebraic equations, I recommend visiting Khan Academy, a fantastic resource offering free courses and practice exercises on a wide range of mathematical topics, from basic arithmetic to advanced calculus. Another excellent source for deeper understanding of number theory principles is Brilliant.org, which provides interactive learning experiences.