Solving: $(3 imes ext{sqrt}{3})^{-x+1} = (1/3) imes 27^{x-5}$

Let's dive into solving this interesting exponential equation step by step. Exponential equations might seem intimidating at first, but with a systematic approach, they can be tackled efficiently. We'll break down the equation $(3 imes ext{sqrt}{3})^{-x+1} = (1/3) imes 27^{x-5}$, making sure each step is clear and easy to follow. Our main goal is to find the value of 'x' that satisfies the equation. We'll achieve this by manipulating the equation, using properties of exponents and logarithms, and simplifying until we isolate 'x'. Remember, the key to solving exponential equations is often to express all terms with the same base. This allows us to equate the exponents and solve for the unknown variable. So, let's put on our math hats and get started!

Breaking Down the Equation

To start, let's rewrite the equation in a more manageable form. The given equation is $(3 imes ext{sqrt}{3})^{-x+1} = (1/3) imes 27^{x-5}$. Our first step involves expressing all terms using a common base. Since 3 is the smallest base present, we'll aim to rewrite everything in terms of base 3. This involves understanding and applying exponent rules effectively. One crucial rule is that $a^{m} imes a^{n} = a^{m+n}$. Another important rule is $(a^{m})^{n} = a^{mn}$. Additionally, we should remember that the square root of a number can be expressed as a fractional exponent (e.g., $ ext{sqrt}{3} = 3^{1/2}$). By using these rules, we can simplify the equation and make it easier to solve. We also need to recall that $1/a = a^{-1}$, which will be useful in rewriting the fraction $1/3$. Let’s begin by rewriting each term individually to ensure clarity and accuracy. This careful approach will lay the foundation for solving the equation effectively.

Rewriting the Left Side

The left side of the equation is $(3 imes ext{sqrt}{3})^{-x+1}$. We can rewrite $ extsqrt}{3}$ as $3^{1/2}$. So, the expression becomes $(3 imes 3^{1/2})^{-x+1}$. Now, using the rule $a^{m} imes a^{n} = a^{m+n}$, we can combine the terms inside the parentheses $3^{1 imes 3^{1/2} = 3^{1 + 1/2} = 3^{3/2}$. Thus, the left side simplifies to $(3^{3/2})^{-x+1}$. Applying the rule $(a^{m})^{n} = a^{mn}$, we get $3^{(3/2)(-x+1)}$. This further simplifies to $3^{(-3/2)x + 3/2}$. By rewriting the left side in this way, we've expressed it as a power of 3, which aligns with our goal of having a common base throughout the equation. This simplification is a crucial step in solving exponential equations, as it allows us to compare exponents directly. Let's keep this simplified form in mind as we move on to rewriting the right side of the equation. Remember, the more we simplify each part, the easier it will be to solve the overall equation. The key is to break down complex expressions into smaller, more manageable components.

Rewriting the Right Side

The right side of the equation is $(1/3) imes 27^{x-5}$. We need to rewrite both terms as powers of 3. First, let's address $1/3$. We can express $1/3$ as $3^{-1}$ using the property $1/a = a^{-1}$. Next, we need to rewrite 27 as a power of 3. We know that $27 = 3^3$, so we can rewrite $27^{x-5}$ as $(3^3)^{x-5}$. Applying the rule $(a^{m})^{n} = a^{mn}$, we get $3^{3(x-5)}$, which simplifies to $3^{3x-15}$. Now, we can rewrite the entire right side as $3^{-1} imes 3^{3x-15}$. Using the rule $a^{m} imes a^{n} = a^{m+n}$, we combine the terms: $3^{-1 + (3x-15)} = 3^{3x-16}$. By rewriting the right side in this form, we have successfully expressed it as a power of 3, just like the left side. This is a critical step because it allows us to equate the exponents and solve for x. The simplification process ensures that we can work with a consistent base, which is essential for solving exponential equations efficiently. With both sides of the equation now expressed as powers of 3, we are well-prepared to move on to the next stage of the solution.

Equating the Exponents

Now that we've rewritten both sides of the equation with a common base of 3, we have: $3^{(-3/2)x + 3/2} = 3^{3x-16}$. Since the bases are equal, we can equate the exponents. This means setting $(-3/2)x + 3/2$ equal to $3x - 16$. This step is crucial because it transforms the exponential equation into a linear equation, which is much easier to solve. By equating the exponents, we eliminate the exponential part of the equation and focus solely on solving for 'x'. This technique is a fundamental aspect of solving exponential equations, and it simplifies the problem significantly. The resulting linear equation can be solved using standard algebraic methods. It's important to ensure that the bases are the same before equating exponents; otherwise, this method cannot be applied. So, let's proceed with solving the linear equation to find the value of 'x'. The next steps will involve isolating 'x' on one side of the equation, using basic algebraic operations such as addition, subtraction, multiplication, and division.

Solving the Linear Equation

We now have the linear equation $(-3/2)x + 3/2 = 3x - 16$. To solve for x, we first want to eliminate the fraction. We can do this by multiplying every term in the equation by 2, which gives us $-3x + 3 = 6x - 32$. Next, we want to get all the 'x' terms on one side of the equation and the constants on the other side. Let's add 3x to both sides: $3 = 9x - 32$. Now, add 32 to both sides: $35 = 9x$. Finally, divide both sides by 9 to isolate x: $x = 35/9$. So, the solution to the equation is $x = 35/9$. This value of x satisfies the original exponential equation. Solving linear equations is a fundamental skill in algebra, and it's essential for solving many types of mathematical problems, including exponential equations. The steps we followed here – eliminating fractions, collecting like terms, and isolating the variable – are standard techniques for solving linear equations. By carefully applying these steps, we can confidently find the value of x that makes the equation true. Now that we have the solution, let's take a moment to summarize our steps and verify our answer.

Verification

To ensure our solution is correct, we should substitute $x = 35/9$ back into the original equation: $(3 imes ext{sqrt}{3})^{-x+1} = (1/3) imes 27^{x-5}$. Substituting $x = 35/9$, we get $(3 imes 3^{1/2})^{-(35/9)+1} = (1/3) imes 27^{(35/9)-5}$. Let's simplify the left side first. We have $(3^{3/2})^{-(35/9)+1} = (3^{3/2})^{-26/9} = 3^{(3/2)(-26/9)} = 3^{-13/3}$. Now let's simplify the right side. We have $(1/3) imes 27^{(35/9)-5} = 3^{-1} imes (3^3)^{(35/9)-5} = 3^{-1} imes 3^{3(35/9 - 5)} = 3^{-1} imes 3^{3(35/9 - 45/9)} = 3^{-1} imes 3^{3(-10/9)} = 3^{-1} imes 3^{-10/3} = 3^{-1 - 10/3} = 3^{-3/3 - 10/3} = 3^{-13/3}$. Since both sides simplify to $3^{-13/3}$, our solution $x = 35/9$ is correct. Verification is a critical step in solving any equation, as it helps to catch any errors made during the solution process. By substituting the solution back into the original equation, we can confirm that it satisfies the equation and that our answer is accurate. This process reinforces our understanding of the equation and the solution, providing confidence in our result. Let's summarize the entire process we followed to solve this exponential equation.

Conclusion

In summary, we solved the exponential equation $(3 imes ext{sqrt}{3})^{-x+1} = (1/3) imes 27^{x-5}$ by first rewriting all terms with a common base of 3. We then equated the exponents, resulting in a linear equation. Solving the linear equation, we found that $x = 35/9$. Finally, we verified our solution by substituting it back into the original equation. This systematic approach to solving exponential equations involves understanding and applying the properties of exponents, simplifying expressions, and using algebraic techniques to isolate the variable. Remember, practice is key to mastering these skills. By working through various examples, you'll become more comfortable with the steps involved and more confident in your ability to solve exponential equations. We encourage you to explore additional resources and practice problems to further enhance your understanding. Understanding these concepts not only helps in mathematics but also in various fields of science and engineering where exponential functions are commonly used. If you're eager to learn more about exponential equations and related topics, consider checking out trusted educational websites like Khan Academy for more in-depth explanations and practice exercises.