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Simplify the expression. Assume \(a, b, c, d>0\) $$\frac{(7 a)^{2}(5 b)^{3 / 2}}{(5 a)^{3 / 2}(7 b)^{4}}$$

Short Answer

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Question: Simplify the expression: $$\frac{(7 a)^{2}(5 b)^{3 / 2}}{(5 a)^{3 / 2}(7 b)^{4}}$$, where \(a, b, c, d > 0\). Answer: $$\frac{a^{\frac{1}{2}}}{49b^{\frac{5}{2}}}$$

Step by step solution

01

Rewrite the expression in terms of powers

We will rewrite the expression in terms of powers such that it becomes easier to simplify. The given expression is: $$\frac{(7 a)^{2}(5 b)^{3 / 2}}{(5 a)^{3 / 2}(7 b)^{4}}$$ Using the power rule (multiplication) \( (x^m)^n = x^{mn},\) we rewrite the expression as: $$\frac{7^{2}a^{2} \cdot 5^{\frac{3} {2}}b^{\frac{3} {2}}}{5^{\frac{3} {2}}a^{\frac{3}{2}} \cdot 7^{4} b^{4}}$$
02

Simplify exponential terms by utilizing exponent rules

We simplify exponential terms using the quotient rule of exponents (division): \(x^m/x^n = x^{m-n}\). Our expression now becomes: $$\frac{7^{2-4}a^{2-\frac{3}{2}} \cdot 5^{\frac{3}{2}-\frac{3}{2}}b^{\frac{3}{2}-4}}{1}$$ $$= 7^{-2}a^{\frac{1}{2}} \cdot 5^{0}b^{-\frac{5}{2}}$$
03

Utilize exponential rules for negative exponents

Next, we use the negative exponent rule: \(x^{-n} = \frac{1}{x^n}\). This gives us: $$\frac{a^{\frac{1}{2}}}{7^2} \cdot \frac{5^0}{b^{\frac{5}{2}}}$$
04

Simplify and write the final expression

Now we simplify the constant terms, noting that \(5^0 = 1\) and combine the remaining terms to get the final simplified expression: $$\frac{a^{\frac{1}{2}}}{49b^{\frac{5}{2}}}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponent Rules
Understanding exponent rules is crucial when simplifying exponential expressions. These rules provide a systematic method to manipulate expressions involving powers. One fundamental rule is the power rule, which allows you to multiply exponents when raising an exponent to another power, as seen with \( (x^m)^n = x^{mn} \). When dividing exponents with the same base, the quotient rule comes into play: \(x^m / x^n = x^{m-n}\), effectively subtracting the exponents. Lastly, the negative exponent rule is particularly helpful to simplify terms with negative powers: \(x^{-n} = 1/x^n\), meaning you take the reciprocal of the base raised to the positive power. In the provided exercise, these rules are applied successively to reach the simplified form of the given expression.

For instance, \(7^{2} / 7^{4}\) becomes \(7^{-2}\) using the quotient rule. Then, according to the negative exponent rule, \(7^{-2}\) transforms into \(1/7^{2}\). This process systematically reduces complex expressions into simpler ones that are easier to understand and calculate.
Radicals and Rational Exponents
Radicals and rational exponents are two sides of the same coin. A radical expression can be rewritten as an expression with a rational exponent and vice versa. This conversion is beneficial for simplification. The \(n\)th root of a number \(x\) is equivalent to raising \(x\) to the power of \(1/n\), symbolized as \(\sqrt[n]{x} = x^{1/n}\).

Conversion between Radicals and Rational Exponents

When simplifying, converting to a rational exponent can make it easier to apply the quotient rule for exponents. In the exercise, \(5b)^{3 / 2}\) represents the square root of \(5b\) cubed, which could also be written as \(\sqrt[2]{(5b)^3}\). Similarly, the term \(a^{\frac{1}{2}}\) signifies the square root of \(a\). These conversions are applied to consolidate terms with fractional exponents, ultimately leading to a more straightforward expression.
Algebraic Simplification
The goal of algebraic simplification is to present expressions in their most reduced and easily interpretable form. After applying the exponent rules and understanding the interplay between radicals and rational exponents, algebraic simplification involves combining like terms and reducing fractions whenever possible. In the provided exercise, this involved recognizing that \(5^{\frac{3}{2} - \frac{3}{2}} = 5^{0} = 1\) and thus eliminating that term from the equation.

Final Steps in Simplification

Towards the end of the process, we also simplified \(7^{2}\) to \(49\) and rewrote \(b^{-\frac{5}{2}}\) as \(1 / b^{\frac{5}{2}}\). By performing these steps, the original complex expression was reduced to \(\frac{a^{\frac{1}{2}}}{49b^{\frac{5}{2}}}\), which is much simpler to interpret. Through algebraic simplification, students can achieve clarity and ease in understanding otherwise daunting expressions.

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Most popular questions from this chapter

The population of India was approximately 1030 million in 2001 and was 967 million in \(1997 .\) If the population continues to grow exponentially at the same rate, what will it be in \(2010 ?\)

According to one theory of learning, the number of words per minute \(N\) that a person can type after \(t\) weeks of practice is given by \(N=c\left(1-e^{-k t}\right),\) where \(c\) is an upper limit that \(N\) cannot exceed and \(k\) is a constant that must be determined experimentally for each person. (a) If a person can type 50 wpm (words per minute) after four weeks of practice and 70 wpm after eight weeks, find the values of \(k\) and \(c\) for this person. According to the theory, this person will never type faster than \(c\) wpm. (b) Another person can type 50 wpm after four weeks of practice and 90 wpm after eight weeks. How many weeks must this person practice to be able to type 125 wpm?

The population of St. Petersburg, Florida (in thousands) can be approximated by the function $$ g(x)=-127.9+81.91 \ln x \quad(x \geq 70) $$ where \(x=70\) corresponds to 1970 (a) Estimate the population in 1995 and 2003 . (b) If this model remains accurate, when will the population be \(260,000 ?\)

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