Question

Of the following four numbers, the largest is:

• A. $3^{210}$
• B. $7^{140}$
• C. ${\left(17\right)}^{105}$
• D. ${\left(31\right)}^{84}$

Answer 1

The correct option is C

${\left(17\right)}^{105}$

Step 1: Compare given numbers by making their powers the same:

$$\begin{gathered} {\left(3\right)}^{210}={\left(3\right)}^{6\times 35}={\left(3^6\right)}^{35}={\left(729\right)}^{35}........\left(1\right) \\ {\left(7\right)}^{140}={\left(7\right)}^{4\times 35}={\left(7^4\right)}^{35}={\left(2401\right)}^{35}.......\left(2\right) \\ {\left(17\right)}^{105}={\left(17\right)}^{3\times 35}={\left({17}^3\right)}^{35}={\left(4913\right)}^{35}.......\left(3\right) \\ \mathrm{Now}{\left(4913\right)}^{35}>{\left(2401\right)}^{35}>{\left(729\right)}^{35} \\ \therefore {\left(17\right)}^{105}>{\left(7\right)}^{140}>{\left(3\right)}^{210}........\left(4\right) \end{gathered}$$

Step 2: Compare the remaining two numbers:

$$\begin{gathered} {\left(17\right)}^{105}={\left(17\right)}^{5\times 21}={\left({17}^5\right)}^{21}={\left(1419857\right)}^{21}........\left(5\right) \\ {\left(31\right)}^{84}={\left(31\right)}^{4\times 21}={\left({31}^4\right)}^{21}={\left(923521\right)}^{21}........\left(6\right) \end{gathered}$$

From $\text{(4) ,(5) and (6)}$, we get,

${\left(17\right)}^{\mathbf{105}}$is the largest value.

Final answer:

Hence, option C is the correct answer.