MHB Find the smallest positive integer n

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To find the smallest positive integer n such that n^16 exceeds 16^18, the calculation involves comparing the two expressions. The value of 16 can be rewritten as 2^4, leading to the inequality n^16 > (2^4)^18, simplifying to n^16 > 2^72. Taking the 16th root, this results in n > 2^(72/16), which simplifies to n > 2^4. Therefore, the smallest integer n that satisfies this condition is 17. The discussion highlights different approaches to arrive at the solution, with some preferring simpler methods.
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Find the smallest positive integer $n $ for which $n^{16}$ exceeds $16^{18}$.
 
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anemone said:
Find the smallest positive integer $n $ for which $n^{16}$ exceeds $16^{18}$.
23 as below

n^16 > 16^18
or n^8 > 16^9 or 4^18
or n^4 > 4^9 or 2^18
or n^2 > 2^9 or 512
n = 22 => n^2 = 484 and n = 23 => n^2 = 529
 
Hello, anemone!

\text{Find the smallest positive integer }n
. . \text{ for which }n^{16}\text{ exceeds }16^{18}.
kaliprasad is correct.
I used a different approach.

We want: .n^{16} \;> \; 16^{18}

. . . . . . . .n^{16} \;>\; (2^4)^{18}

. . . . . . . .n^{16} \;>\;2^{72}

. . . . . . . . . n \;>\;2^{\frac{72}{16}}\;=\;2^{\frac{9}{2}}

. . . . . . . . . n \;>\; 2^{4+\frac{1}{2}} \;=\;2^4 \cdot 2^{\frac{1}{2}}

. . . . . . . . . n \;>\; 16\sqrt{2} \;=\; 22.627417

Therefore: . n \;=\;23

 
soroban said:
Hello, anemone!


kaliprasad is correct.
I used a different approach.

We want: .n^{16} \;> \; 16^{18}

. . . . . . . .n^{16} \;>\; (2^4)^{18}

. . . . . . . .n^{16} \;>\;2^{72}

. . . . . . . . . n \;>\;2^{\frac{72}{16}}\;=\;2^{\frac{9}{2}}

. . . . . . . . . n \;>\; 2^{4+\frac{1}{2}} \;=\;2^4 \cdot 2^{\frac{1}{2}}

. . . . . . . . . n \;>\; 16\sqrt{2} \;=\; 22.627417

Therefore: . n \;=\;23

above approach is more straight forward
 
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