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Q.

Arrange the following in ascending order $3^{34}, 2^{51}, 7^{17}$ , we get

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a

$2^{51} > 3^{34} > 7^{17}$

b

$7^{17} > 2^{51} > 3^{34}$

c

$3^{34} > 7^{17} > 2^{51}$

d

$3^{34} > 2^{51} > 7^{17}$

answer is A.

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Detailed Solution

$\begin{array}{r} 3^{34} = {(3^{2})}^{17} = 9^{17} \\ 2^{51} = {(2^{3})}^{17} = 8^{17} \end{array}$

Clearly, $9^{17} > 8^{17} > 7^{17}$

or

$9^{17} > 2^{51} > 3^{17} \Rightarrow 3^{34} > 2^{51} > 7^{17}$

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