Simplify The Following Expression:$\left(32 X^{25} Y^{-10}\right)^{-\frac{3}{5}}$

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Introduction

In mathematics, simplifying expressions is a crucial skill that helps us solve problems efficiently. When dealing with exponents, it's essential to understand the rules of exponentiation to simplify complex expressions. In this article, we will focus on simplifying the given expression (32x25yβˆ’10)βˆ’35\left(32 x^{25} y^{-10}\right)^{-\frac{3}{5}} using the rules of exponentiation.

Understanding Exponents

Before we dive into simplifying the expression, let's review the basics of exponents. An exponent is a small number that is written to the right of a base number and indicates how many times the base number should be multiplied by itself. For example, in the expression x3x^3, the exponent 3 indicates that the base number xx should be multiplied by itself three times: xβ‹…xβ‹…xx \cdot x \cdot x.

Simplifying the Expression

To simplify the given expression, we need to apply the rules of exponentiation. The expression (32x25yβˆ’10)βˆ’35\left(32 x^{25} y^{-10}\right)^{-\frac{3}{5}} can be simplified using the following steps:

Step 1: Simplify the Coefficient

The coefficient of the expression is 32. To simplify the coefficient, we can rewrite it as 252^5. This is because 32 can be expressed as 252^5.

Step 2: Simplify the Exponents

Now, let's focus on simplifying the exponents. The expression contains two variables: xx and yy. The exponent of xx is 25, and the exponent of yy is -10. To simplify the exponents, we need to apply the rule of exponentiation that states (am)n=amβ‹…n(a^m)^n = a^{m \cdot n}.

Step 3: Apply the Rule of Exponentiation

Using the rule of exponentiation, we can simplify the expression as follows:

(32x25yβˆ’10)βˆ’35=(25x25yβˆ’10)βˆ’35\left(32 x^{25} y^{-10}\right)^{-\frac{3}{5}} = \left(2^5 x^{25} y^{-10}\right)^{-\frac{3}{5}}

=2βˆ’5β‹…35xβˆ’25β‹…35yβˆ’10β‹…βˆ’35= 2^{-5 \cdot \frac{3}{5}} x^{-25 \cdot \frac{3}{5}} y^{-10 \cdot -\frac{3}{5}}

=2βˆ’3xβˆ’15y6= 2^{-3} x^{-15} y^{6}

Step 4: Simplify the Expression Further

Now that we have simplified the exponents, we can simplify the expression further by rewriting the coefficients and variables in their simplest form.

2βˆ’3=123=182^{-3} = \frac{1}{2^3} = \frac{1}{8}

xβˆ’15=1x15x^{-15} = \frac{1}{x^{15}}

y6=y6y^{6} = y^6

Therefore, the simplified expression is:

18β‹…1x15β‹…y6\frac{1}{8} \cdot \frac{1}{x^{15}} \cdot y^6

Conclusion

In this article, we simplified the expression (32x25yβˆ’10)βˆ’35\left(32 x^{25} y^{-10}\right)^{-\frac{3}{5}} using the rules of exponentiation. We applied the rule of exponentiation to simplify the exponents and then simplified the expression further by rewriting the coefficients and variables in their simplest form. The simplified expression is 18β‹…1x15β‹…y6\frac{1}{8} \cdot \frac{1}{x^{15}} \cdot y^6.

Frequently Asked Questions

  • What is the rule of exponentiation? The rule of exponentiation states that (am)n=amβ‹…n(a^m)^n = a^{m \cdot n}.
  • How do you simplify exponents? To simplify exponents, you need to apply the rule of exponentiation and then simplify the expression further by rewriting the coefficients and variables in their simplest form.
  • What is the simplified expression of (32x25yβˆ’10)βˆ’35\left(32 x^{25} y^{-10}\right)^{-\frac{3}{5}}? The simplified expression is 18β‹…1x15β‹…y6\frac{1}{8} \cdot \frac{1}{x^{15}} \cdot y^6.

Final Answer

The final answer is 18β‹…1x15β‹…y6\boxed{\frac{1}{8} \cdot \frac{1}{x^{15}} \cdot y^6}.

Q&A: Simplifying Exponents

In the previous article, we simplified the expression (32x25yβˆ’10)βˆ’35\left(32 x^{25} y^{-10}\right)^{-\frac{3}{5}} using the rules of exponentiation. In this article, we will answer some frequently asked questions related to simplifying exponents.

Q1: What is the rule of exponentiation?

A1: The rule of exponentiation states that (am)n=amβ‹…n(a^m)^n = a^{m \cdot n}. This rule helps us simplify exponents by multiplying the exponents together.

Q2: How do you simplify exponents?

A2: To simplify exponents, you need to apply the rule of exponentiation and then simplify the expression further by rewriting the coefficients and variables in their simplest form.

Q3: What is the difference between a positive exponent and a negative exponent?

A3: A positive exponent indicates that the base number should be multiplied by itself a certain number of times. A negative exponent indicates that the base number should be divided by itself a certain number of times.

Q4: How do you simplify an expression with a negative exponent?

A4: To simplify an expression with a negative exponent, you can rewrite the expression with a positive exponent by taking the reciprocal of the base number. For example, xβˆ’3=1x3x^{-3} = \frac{1}{x^3}.

Q5: What is the simplified expression of (32x25yβˆ’10)βˆ’35\left(32 x^{25} y^{-10}\right)^{-\frac{3}{5}}?

A5: The simplified expression is 18β‹…1x15β‹…y6\frac{1}{8} \cdot \frac{1}{x^{15}} \cdot y^6.

Q6: How do you simplify an expression with multiple variables?

A6: To simplify an expression with multiple variables, you need to apply the rule of exponentiation to each variable separately and then simplify the expression further by rewriting the coefficients and variables in their simplest form.

Q7: What is the rule for multiplying exponents with the same base?

A7: The rule for multiplying exponents with the same base is amβ‹…an=am+na^m \cdot a^n = a^{m + n}. This rule helps us simplify expressions by adding the exponents together.

Q8: How do you simplify an expression with a fraction as an exponent?

A8: To simplify an expression with a fraction as an exponent, you can rewrite the fraction as a decimal or a percentage and then apply the rule of exponentiation.

Q9: What is the simplified expression of (12x3y4)βˆ’2\left(\frac{1}{2} x^3 y^4\right)^{-2}?

A9: To simplify this expression, we need to apply the rule of exponentiation and then simplify the expression further by rewriting the coefficients and variables in their simplest form.

(12x3y4)βˆ’2=(12)βˆ’2β‹…(x3)βˆ’2β‹…(y4)βˆ’2\left(\frac{1}{2} x^3 y^4\right)^{-2} = \left(\frac{1}{2}\right)^{-2} \cdot \left(x^3\right)^{-2} \cdot \left(y^4\right)^{-2}

=22β‹…1x6β‹…1y8= 2^2 \cdot \frac{1}{x^6} \cdot \frac{1}{y^8}

=4β‹…1x6β‹…1y8= 4 \cdot \frac{1}{x^6} \cdot \frac{1}{y^8}

Q10: How do you simplify an expression with a variable as an exponent?

A10: To simplify an expression with a variable as an exponent, you need to apply the rule of exponentiation and then simplify the expression further by rewriting the coefficients and variables in their simplest form.

Conclusion

In this article, we answered some frequently asked questions related to simplifying exponents. We covered topics such as the rule of exponentiation, simplifying expressions with negative exponents, and simplifying expressions with multiple variables. We also provided examples to illustrate the concepts and rules discussed in the article.

Final Answer

The final answer is 18β‹…1x15β‹…y6\boxed{\frac{1}{8} \cdot \frac{1}{x^{15}} \cdot y^6}.