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Fermat
Fermat
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Hint write $9=7\times 1.289$ then $9^{17}={7\times 1.28)^{17}$. Now divide it by $7^{19}$ you get $\frac{{1.28}^{17}}{49}$ .Now using binomial for $(1+0.28)^{17}$ and writing terms till $0.28^4$ we get value approximately as $66$ so $9^{17}>7^{19}$$9^{17}=7\times 1.28^{17}$. Now divide it by $7^{19}$ you get $\frac{{1.28}^{17}}{49}$ .Now using binomial for $(1+0.28)^{17}$ and writing terms till $0.28^4$ we get value approximately as $66$ so $9^{17}>7^{19}$ note that according to me there will be always some calculation as it's a question on number theory.

Hint write $9=7\times 1.289$ then $9^{17}={7\times 1.28)^{17}$. Now divide it by $7^{19}$ you get $\frac{{1.28}^{17}}{49}$ .Now using binomial for $(1+0.28)^{17}$ and writing terms till $0.28^4$ we get value approximately as $66$ so $9^{17}>7^{19}$ note that according to me there will be always some calculation as it's a question on number theory.

Hint write $9=7\times 1.289$ then $9^{17}=7\times 1.28^{17}$. Now divide it by $7^{19}$ you get $\frac{{1.28}^{17}}{49}$ .Now using binomial for $(1+0.28)^{17}$ and writing terms till $0.28^4$ we get value approximately as $66$ so $9^{17}>7^{19}$ note that according to me there will be always some calculation as it's a question on number theory.

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Archis Welankar
Archis Welankar
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Hint write $9=7\times 1.289$ then $9^{17}={7\times 1.28)^{17}$. Now divide it by $7^{19}$ you get $\frac{{1.28}^{17}}{49}$ .Now using binomial for $(1+0.28)^{17}$ and writing terms till $0.28^4$ we get value approximately as $66$ so $9^{17}>7^{19}$ note that according to me there will be always some calculation as it's a question on number theory.