Solving For Variables In Simplified Expressions A Step-by-Step Guide

by Chloe Fitzgerald 69 views

Hey guys! Today, we're diving into a super fun math problem that involves simplifying expressions with exponents. It might look a little intimidating at first, but trust me, we'll break it down step by step so it's totally easy to understand. We're going to solve for the variables a, b, and c in this expression:

25â‹…8416=25â‹…(2a)424=25â‹…2b24=2c\frac{2^5 \cdot 8^4}{16}=\frac{2^5 \cdot(2^a)^4}{2^4}=\frac{2^5 \cdot 2^b}{2^4}=2^c

Ready to get started? Let's jump right in!

Breaking Down the Expression

To solve this expression, we need to understand the properties of exponents. Remember, exponents tell us how many times to multiply a number by itself. For example, 2^5 means 2 multiplied by itself 5 times (2 * 2 * 2 * 2 * 2 = 32). Also, keep in mind that 8 and 16 can be expressed as powers of 2, which will help us simplify the expression. Our main goal here is to find the values of a, b, and c that make the equation true.

The expression we're working with is:

25â‹…8416=25â‹…(2a)424=25â‹…2b24=2c\frac{2^5 \cdot 8^4}{16}=\frac{2^5 \cdot(2^a)^4}{2^4}=\frac{2^5 \cdot 2^b}{2^4}=2^c

Let's tackle each part of this equation one at a time. We'll start by rewriting the numbers in terms of powers of 2. This will make it much easier to apply the exponent rules and simplify the expression. Remember, the key to solving these kinds of problems is to take it slow and break it down into manageable steps. Don't rush, and make sure you understand each step before moving on to the next. By doing this, you'll not only solve the problem correctly but also gain a deeper understanding of how exponents work.

Rewriting 8 and 16 as Powers of 2

First things first, we need to express 8 and 16 as powers of 2. This is a crucial step because it allows us to use the properties of exponents to simplify the expression. We know that:

  • 8 = 2 * 2 * 2 = 2^3
  • 16 = 2 * 2 * 2 * 2 = 2^4

Now that we have these values, we can substitute them back into our original expression. This substitution will make the entire expression much easier to work with. So, let's replace 8 with 2^3 and 16 with 2^4 in the original equation. This step is all about making the expression look simpler and more manageable. Once we've done this, we can start applying the exponent rules to further simplify it.

Substituting these values into the original expression, we get:

25â‹…(23)424=25â‹…2b24=2c\frac{2^5 \cdot (2^3)^4}{2^4}=\frac{2^5 \cdot 2^b}{2^4}=2^c

Notice how we've replaced 8 with 2^3, and we've kept the rest of the equation the same. Now, we're ready to move on to the next step, which involves simplifying the expression further using the power of a power rule.

Solving for Variable a

Alright, let's solve for a! We're focusing on this part of the equation:

25â‹…(2a)424\frac{2^5 \cdot (2^a)^4}{2^4}

The first thing we need to do is simplify the term (2a)4. Remember the power of a power rule? It says that (xm)n = x^(mn). This rule is super handy for simplifying expressions like this. So, when we apply this rule to (2a)4, we get 2^(a4) or simply 2^(4a).

Now, let's rewrite our expression with this simplification:

25â‹…24a24\frac{2^5 \cdot 2^{4a}}{2^4}

Now, let's look back at the original equation. We can see that:

25â‹…(2a)424=25â‹…(23)424\frac{2^5 \cdot (2^a)^4}{2^4} = \frac{2^5 \cdot (2^3)^4}{2^4}

We already figured out that 8 is 2^3, so we substituted that into the equation. Now, we can equate the exponents in the numerators:

(2a)4=(23)4(2^a)^4 = (2^3)^4

Using the power of a power rule again, we get:

24a=23â‹…42^{4a} = 2^{3 \cdot 4}

24a=2122^{4a} = 2^{12}

Since the bases are the same (both are 2), we can set the exponents equal to each other:

4a=124a = 12

Now, we just need to solve for a. To do that, we divide both sides of the equation by 4:

a=124a = \frac{12}{4}

a=3a = 3

Tada! We've found the value of a. It's 3. This means that when we replace a with 3 in the original expression, everything will balance out. This is a great feeling, right? We've taken a big step towards solving the whole problem. Now, let's move on to the next variable.

So, we've successfully navigated the power of a power rule and solved for a. Give yourself a pat on the back! Understanding and applying these exponent rules is key to mastering these types of problems. With each step we take, the bigger picture becomes clearer, and we're one step closer to cracking the entire problem.

Solving for Variable b

Awesome, let's keep the momentum going and solve for b! We'll be working with this part of the equation:

25â‹…(2a)424=25â‹…2b24\frac{2^5 \cdot (2^a)^4}{2^4} = \frac{2^5 \cdot 2^b}{2^4}

We already know that a = 3, so let's substitute that value back into the equation:

25â‹…(23)424=25â‹…2b24\frac{2^5 \cdot (2^3)^4}{2^4} = \frac{2^5 \cdot 2^b}{2^4}

Now, let's simplify the left side of the equation first. We already know that (23)4 is 2^(3*4) which equals 2^12. So, we can rewrite the equation as:

25â‹…21224=25â‹…2b24\frac{2^5 \cdot 2^{12}}{2^4} = \frac{2^5 \cdot 2^b}{2^4}

Next, we can use the product of powers rule, which says that x^m * x^n = x^(m+n). This rule is super useful for combining terms with the same base. Applying this rule to the numerator on both sides, we get:

25+1224=25+b24\frac{2^{5+12}}{2^4} = \frac{2^{5+b}}{2^4}

21724=25+b24\frac{2^{17}}{2^4} = \frac{2^{5+b}}{2^4}

Now, we have the same denominator on both sides of the equation, which is great! This means we can focus on the numerators. For the two fractions to be equal, their numerators must be equal as well. So, we can set the exponents in the numerators equal to each other:

217=25+b2^{17} = 2^{5+b}

Since the bases are the same, we can equate the exponents:

17=5+b17 = 5 + b

Now, we just need to solve for b. To do that, we subtract 5 from both sides of the equation:

17−5=b17 - 5 = b

12=b12 = b

So, we've found the value of b! It's 12. This means that when we substitute b with 12 in the equation, both sides will be equal. We're on a roll, guys! We've solved for a and b. Now, there's just one more variable to go.

Solving for b involved using the product of powers rule and then equating the exponents. These are essential techniques for simplifying expressions with exponents, so make sure you're comfortable with them. We're building a solid foundation for tackling more complex problems in the future. Let's keep this momentum going and solve for the final variable!

Solving for Variable c

Alright, let's wrap this up by solving for c! We're now focusing on the final part of the equation:

25â‹…2b24=2c\frac{2^5 \cdot 2^b}{2^4} = 2^c

We already know that b = 12, so let's substitute that value into the equation:

25â‹…21224=2c\frac{2^5 \cdot 2^{12}}{2^4} = 2^c

Just like before, we can use the product of powers rule to simplify the numerator. Remember, x^m * x^n = x^(m+n). So, we can combine the terms in the numerator:

25+1224=2c\frac{2^{5+12}}{2^4} = 2^c

21724=2c\frac{2^{17}}{2^4} = 2^c

Now, we need to simplify the left side of the equation further. We can use the quotient of powers rule, which says that x^m / x^n = x^(m-n). This rule is super helpful for dividing terms with the same base. Applying this rule, we get:

217−4=2c2^{17-4} = 2^c

213=2c2^{13} = 2^c

Now, we have a very simple equation. We have the same base on both sides, so we can simply equate the exponents:

13=c13 = c

And there you have it! We've found the value of c. It's 13. This completes our journey of solving for all the variables in the expression. We've successfully navigated the world of exponents and come out victorious!

Solving for c involved using both the product of powers rule and the quotient of powers rule. These rules are fundamental to simplifying expressions with exponents, so make sure you have a good grasp of them. We've tackled each part of the problem step by step, and now we have a complete solution.

Final Answer

Woo-hoo! We did it! We've successfully solved for all the variables in the expression. Let's recap our answers:

  • a = 3
  • b = 12
  • c = 13

So, the final answers are:

a=3b=12c=13\begin{array}{l} a = 3 \\ b = 12 \\ c = 13 \end{array}

We started with a seemingly complex expression, but by breaking it down into smaller, manageable steps and applying the rules of exponents, we were able to find the values of a, b, and c. This is a testament to the power of methodical problem-solving. Remember, guys, math is like a puzzle, and each step we take is like fitting a piece into place. The more pieces we fit, the clearer the picture becomes.

We've covered a lot in this article, from rewriting numbers as powers of 2 to applying the product of powers and quotient of powers rules. These are essential skills for simplifying expressions and solving equations with exponents. Keep practicing these techniques, and you'll become a pro in no time!

I hope you found this explanation helpful and that you now have a better understanding of how to solve these types of problems. Keep practicing, and you'll become a math whiz in no time! If you have any questions or want to explore more math problems, feel free to ask. Keep up the great work, guys!