Question

which expression is equivalent to \\((15a^{0}b^{2}c^{34})(3a^{16}b^{-29}c^{0})\\) for all values of \\(a\\), \\(b\\), and \\(c\\) where the expression is defined?\
a \\(\frac{18}{b^{58}}\\)\
b \\(\frac{45}{b^{58}}\\)\
c \\(\frac{18a^{16}c^{34}}{b^{27}}\\)\
d \\(\frac{45a^{16}c^{34}}{b^{27}}\\)\
\\(\bigcirc\\) a\
\\(\bigcirc\\) d\
\\(\bigcirc\\) c\
\\(\bigcirc\\) b

Explanation:

Step1: Multiply constant coefficients

$15 \times 3 = 45$

Step2: Simplify $a$-terms (use $x^m \cdot x^n=x^{m+n}$)

$a^0 \cdot a^{16} = a^{0+16} = a^{16}$

Step3: Simplify $b$-terms (use $x^m \cdot x^n=x^{m+n}$)

$b^2 \cdot b^{-29} = b^{2+(-29)} = b^{-27} = \frac{1}{b^{27}}$

Step4: Simplify $c$-terms (use $x^m \cdot x^n=x^{m+n}$)

$c^{-34} \cdot c^0 = c^{-34+0} = c^{-34} = \frac{1}{c^{34}}$

Step5: Combine all simplified terms

$45 \times a^{16} \times \frac{1}{b^{27}} \times \frac{1}{c^{34}} = \frac{45a^{16}}{b^{27}c^{34}}$
(Note: This can also be written as $\frac{45a^{16}c^{-34}}{b^{27}}$, which matches the structure of option D when recognizing negative exponents move to the denominator)

Answer:

D. $\frac{45a^{16}c^{34}}{b^{27}}$