Full solution

(b d)^{\frac{1}{4}} \cdot\left(\frac{b}{d}\right)^{4}

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Which of the following expressions is equivalent to the given expression assuming

d

is nonzero?

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Choose

1

answer:

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(A)

b^{\frac{17}{4}} d^{\frac{17}{4}}

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(B)

b^{\frac{17}{4}} d^{\frac{15}{4}}

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(C)

b^{2}

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(D)

\frac{b}{d}

Simplify Term $1$ : Simplify the expression $(bd)^{\frac{1}{4}}\cdot\left(\frac{b}{d}\right)^{4}$ . We have two terms: $(bd)^{\frac{1}{4}}$ and $\left(\frac{b}{d}\right)^{4}$ . We will simplify each term separately before combining them.
Simplify Term $2$ : Simplify the first term $(bd)^{\frac{1}{4}}$ . Since the exponent is $\frac{1}{4}$ , we are essentially taking the fourth root of $bd$ . This term remains as is for now.
Combine Terms: Simplify the second term $\left(\frac{b}{d}\right)^{4}$ . When we raise a fraction to an exponent, we raise both the numerator and the denominator to that exponent. So, $\left(\frac{b}{d}\right)^{4}$ becomes $\frac{b^4}{d^4}$ .
Multiply Numerators: Combine the simplified terms. $\newline$ We now have $(bd)^{\frac{1}{4}} \times \left(\frac{b^4}{d^4}\right)$ . To combine these, we multiply the numerators and the denominators separately.
Realize Mistake: Multiply the numerators and the denominators. $\newline$ Multiplying the numerators: $(b \times b^4) = b^{(1+4)} = b^5$ . $\newline$ Multiplying the denominators: $(d^{(1/4)} \times d^4) = d^{(1/4 + 4)} = d^{(17/4)}$ . $\newline$ So, the combined expression is $\frac{b^5}{d^{(17/4)}}$ .
Realize Mistake: Multiply the numerators and the denominators. $\newline$ Multiplying the numerators: $(b \times b^4) = b^{(1+4)} = b^5$ . $\newline$ Multiplying the denominators: $(d^{(1/4)} \times d^4) = d^{(1/4 + 4)} = d^{(17/4)}$ . $\newline$ So, the combined expression is $\frac{b^5}{d^{(17/4)}}$ .Realize the mistake in the previous step and correct it. $\newline$ We made a mistake in the previous step. We should have applied the property of exponents correctly when multiplying terms with the same base. The correct way to combine the terms is to add the exponents for the base $b$ and subtract the exponents for the base $d$ , since one $d$ is in the denominator.