Prove And Simplify Exponential Expressions Q187 And Q188

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Understanding the Problem

This question involves proving a mathematical identity related to exponents and fractions. The core task is to manipulate the given expression using the laws of exponents to show that it simplifies to 1. We will delve into each step, providing clear explanations and justifications for the mathematical operations performed. Understanding these principles is fundamental not only for solving this particular problem but also for tackling more complex algebraic manipulations.

The initial expression is a product of three terms, each involving a fraction raised to a fractional power. The variables a{a}, b{b}, and c{c} represent exponents, and x{x} is the base. The condition x≠0{x \neq 0} is crucial because division by zero is undefined in mathematics. This constraint ensures the expression is mathematically valid. The goal is to apply exponent rules such as the quotient rule and the power of a power rule to simplify the expression. By breaking down each term and applying these rules systematically, we aim to demonstrate that the product of these terms indeed equals 1.

Detailed Solution

To prove the given identity, we start by applying the quotient rule of exponents, which states that xmxn=xm−n{\frac{x^m}{x^n} = x^{m-n}}. This rule is foundational to simplifying expressions involving division of powers with the same base. We will apply this rule to each term within the product, which will allow us to rewrite the expression in a more manageable form. The application of the quotient rule is a key step in untangling the complexity of the original expression and sets the stage for further simplification using additional exponent rules.

Starting with the first term, (xaxb)1a+b{\left(\frac{x^a}{x^b}\right)^{\frac{1}{a+b}}} can be rewritten as (xa−b)1a+b{(x^{a-b})^{\frac{1}{a+b}}}. Similarly, the second term, (xbxc)1b+c{\left(\frac{x^b}{x^c}\right)^{\frac{1}{b+c}}} becomes (xb−c)1b+c{(x^{b-c})^{\frac{1}{b+c}}}, and the third term, (xcxa)1c+a{\left(\frac{x^c}{x^a}\right)^{\frac{1}{c+a}}} is transformed into (xc−a)1c+a{(x^{c-a})^{\frac{1}{c+a}}}. These transformations are direct applications of the quotient rule, making the expression more amenable to further simplification. The next step involves using the power of a power rule to handle the exponents outside the parentheses.

Next, we use the power of a power rule, which states that (xm)n=xmn{(x^m)^n = x^{mn}}. Applying this rule to each term, we get:

  • (xa−b)1a+b=xa−ba+b{(x^{a-b})^{\frac{1}{a+b}} = x^{\frac{a-b}{a+b}}}
  • (xb−c)1b+c=xb−cb+c{(x^{b-c})^{\frac{1}{b+c}} = x^{\frac{b-c}{b+c}}}
  • (xc−a)1c+a=xc−ac+a{(x^{c-a})^{\frac{1}{c+a}} = x^{\frac{c-a}{c+a}}}

This step significantly simplifies the original terms by combining the exponents. The power of a power rule allows us to multiply the exponents, which results in single fractional exponents for each term. This simplification is crucial for the next step, where we will combine these terms using the product rule of exponents. The expression is now in a form where the product rule can be effectively applied.

Now, the original expression becomes:

xa−ba+b⋅xb−cb+c⋅xc−ac+a{x^{\frac{a-b}{a+b}} \cdot x^{\frac{b-c}{b+c}} \cdot x^{\frac{c-a}{c+a}}}

When multiplying powers with the same base, we add the exponents. This is another fundamental rule of exponents known as the product rule, which states that xmâ‹…xn=xm+n{x^m \cdot x^n = x^{m+n}}. Applying this rule allows us to combine the three terms into a single exponential term. The addition of the exponents is the next key step in simplifying the expression and bringing us closer to the final result.

So, we add the exponents:

xa−ba+b+b−cb+c+c−ac+a{x^{\frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{c+a}}}

Now, we need to show that the sum of the exponents is equal to 0. This involves algebraic manipulation of the fractions in the exponent. The exponents must be added carefully, ensuring that we find a common denominator and correctly combine the numerators. This step requires meticulous attention to detail to avoid algebraic errors.

The exponent is:

a−ba+b+b−cb+c+c−ac+a{\frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{c+a}}

To simplify this expression, we need to find a common denominator and add the fractions. The common denominator would be the product of the individual denominators, which is (a+b)(b+c)(c+a){(a+b)(b+c)(c+a)}. The process of combining these fractions involves multiplying each fraction by a form of 1 that will give it the common denominator. This is a standard technique in algebraic manipulation and is crucial for simplifying complex fractional expressions. Once the fractions have a common denominator, we can add their numerators.

Adding these fractions requires careful expansion and simplification. We'll multiply each fraction by the appropriate factors to get the common denominator and then combine the numerators.

The common denominator is (a+b)(b+c)(c+a){(a+b)(b+c)(c+a)}. So, we rewrite the fractions:

(a−b)(b+c)(c+a)+(b−c)(a+b)(c+a)+(c−a)(a+b)(b+c)(a+b)(b+c)(c+a){\frac{(a-b)(b+c)(c+a) + (b-c)(a+b)(c+a) + (c-a)(a+b)(b+c)}{(a+b)(b+c)(c+a)}}

Now, we expand the numerators:

First term:

(a−b)(b+c)(c+a)=(a−b)(bc+ba+c2+ca)=abc+a2b+ac2+a2c−b2c−ab2−bc2−abc=a2b+ac2+a2c−b2c−ab2−bc2{(a-b)(b+c)(c+a) = (a-b)(bc + ba + c^2 + ca) = abc + a^2b + ac^2 + a^2c - b^2c - ab^2 - bc^2 - abc = a^2b + ac^2 + a^2c - b^2c - ab^2 - bc^2}

Second term:

(b−c)(a+b)(c+a)=(b−c)(ac+a2+bc+ba)=abc+a2b+b2c+ab2−ac2−a2c−bc2−abc=a2b+b2c+ab2−ac2−a2c−bc2{(b-c)(a+b)(c+a) = (b-c)(ac + a^2 + bc + ba) = abc + a^2b + b^2c + ab^2 - ac^2 - a^2c - bc^2 - abc = a^2b + b^2c + ab^2 - ac^2 - a^2c - bc^2}

Third term:

(c−a)(a+b)(b+c)=(c−a)(ab+b2+a2+bc)=abc+b2c+a2c+bc2−a2b−ab2−a3−abc=b2c+a2c+bc2−a2b−ab2−a3{(c-a)(a+b)(b+c) = (c-a)(ab + b^2 + a^2 + bc) = abc + b^2c + a^2c + bc^2 - a^2b - ab^2 - a^3 - abc = b^2c + a^2c + bc^2 - a^2b - ab^2 - a^3}

Now, we add the expanded numerators:

(a2b+ac2+a2c−b2c−ab2−bc2)+(a2b+b2c+ab2−ac2−a2c−bc2)+(bc2+a2c+b2c−a2b−ab2−bc2){(a^2b + ac^2 + a^2c - b^2c - ab^2 - bc^2) + (a^2b + b^2c + ab^2 - ac^2 - a^2c - bc^2) + (bc^2 + a^2c + b^2c - a^2b - ab^2 - bc^2)}

This step involves carefully adding the expanded terms, which can be prone to errors if not done meticulously. We need to combine like terms and ensure that all terms are accounted for. The addition of the numerators is a critical step in simplifying the exponent and demonstrating that it indeed equals zero.

Combining like terms:

a2b−a2b+a2b+ac2−ac2+a2c−a2c+a2c−b2c+b2c+b2c−ab2+ab2−ab2−bc2−bc2+bc2=0{a^2b - a^2b + a^2b + ac^2 - ac^2 + a^2c - a^2c + a^2c - b^2c + b^2c + b^2c - ab^2 + ab^2 - ab^2 - bc^2 - bc^2 + bc^2 = 0}

The simplification of the sum of the numerators is a crucial step, as it shows that the exponent ultimately becomes zero. The cancellation of terms is a clear indicator that the sum is zero, which is a necessary condition to prove the original identity. The meticulous process of expanding and combining like terms is what leads us to this result.

Thus, the sum of the exponents is 0, and we have:

x0(a+b)(b+c)(c+a)=x0{x^{\frac{0}{(a+b)(b+c)(c+a)}} = x^0}

Finally, we know that any non-zero number raised to the power of 0 is 1. This is a fundamental property of exponents and is the final step in proving the given identity. This principle is crucial for concluding that the original expression indeed simplifies to 1.

Since x≠0{x \neq 0}, we have:

x0=1{x^0 = 1}

Therefore, we have proven that:

(xaxb)1a+bâ‹…(xbxc)1b+câ‹…(xcxa)1c+a=1{ \left(\frac{x^a}{x^b}\right)^{\frac{1}{a+b}} \cdot \left(\frac{x^b}{x^c}\right)^{\frac{1}{b+c}} \cdot \left(\frac{x^c}{x^a}\right)^{\frac{1}{c+a}} = 1 }

Conclusion

The proof involves the systematic application of exponent rules, including the quotient rule, the power of a power rule, and the product rule. By breaking down the problem into smaller steps and applying these rules sequentially, we demonstrated that the given expression simplifies to 1. This result highlights the power and elegance of exponent rules in simplifying complex mathematical expressions. Understanding and applying these rules effectively is essential for success in algebra and related fields.

(i) (128)27{(128)^{\frac{2}{7}}}

Understanding the Problem

This problem requires simplifying an expression involving a fractional exponent. The key here is to recognize that a fractional exponent can be interpreted as both a power and a root. For example, xmn{x^{\frac{m}{n}}} is equivalent to the n{n}-th root of xm{x^m}, or xmn{\sqrt[n]{x^m}}. Understanding this relationship is crucial for simplifying expressions with fractional exponents. We will also need to express the base, 128 in this case, as a power of a prime number to further simplify the expression. This involves prime factorization, which is a fundamental concept in number theory.

By recognizing the fractional exponent as a combination of a power and a root, we can strategically simplify the expression. The goal is to find a way to eliminate the fractional exponent and express the result as an integer or a simpler form. The prime factorization of the base will help us identify the appropriate root and power, leading to the simplification of the expression.

Detailed Solution

First, we express 128 as a power of 2. Recognizing that 128 is a power of 2 is crucial for simplifying the expression. This is because the denominator of the fractional exponent (7 in this case) hints at taking a seventh root. Thus, expressing 128 as a power of 2 is a strategic step towards simplifying the expression. The prime factorization of 128 allows us to rewrite it in a form that is compatible with the fractional exponent.

We know that 128=27{128 = 2^7}. This is a key observation as it aligns with the denominator of the fractional exponent, 7. The fact that 128 can be expressed as 27{2^7} is critical for simplifying the expression because it allows us to directly apply the properties of exponents. This transformation sets the stage for easily simplifying the expression using the power of a power rule.

So, we can rewrite the expression as:

(27)27{(2^7)^{\frac{2}{7}}}

Now, we apply the power of a power rule, which states that (xm)n=xmn{(x^m)^n = x^{mn}}. This rule is fundamental in simplifying expressions involving exponents and allows us to multiply the exponents together. The application of this rule is a direct and efficient way to simplify the expression by eliminating the outer exponent. By multiplying the exponents, we can eliminate the fractional exponent and express the result in a simpler form.

Applying the rule, we get:

27â‹…27{2^{7 \cdot \frac{2}{7}}}

Simplifying the exponent:

27â‹…27=22{2^{7 \cdot \frac{2}{7}} = 2^2}

The simplification of the exponent is a crucial step that leads to the final answer. By multiplying the exponents and canceling the common factor of 7, we are left with a simple integer exponent. This simplification is a direct result of the power of a power rule and allows us to easily compute the final value of the expression. The simplified exponent indicates that we now need to compute a simple power of 2.

Finally, we compute 22{2^2}:

22=4{2^2 = 4}

Thus, the simplified form of (128)27{(128)^{\frac{2}{7}}} is 4.

Conclusion

Simplifying expressions with fractional exponents involves recognizing the relationship between roots and powers, as well as applying exponent rules such as the power of a power rule. In this case, by expressing 128 as a power of 2 and applying the power of a power rule, we simplified the expression to 4. This highlights the importance of understanding exponent rules and prime factorization in simplifying mathematical expressions. The systematic application of these principles allows us to break down complex expressions into simpler forms.