Find the smallest positive integer n

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anemone
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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!

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

We want: .[tex]n^{16} \;> \; 16^{18}[/tex]

. . . . . . . .[tex]n^{16} \;>\; (2^4)^{18}[/tex]

. . . . . . . .[tex]n^{16} \;>\;2^{72}[/tex]

. . . . . . . . . [tex]n \;>\;2^{\frac{72}{16}}\;=\;2^{\frac{9}{2}}[/tex]

. . . . . . . . . [tex]n \;>\; 2^{4+\frac{1}{2}} \;=\;2^4 \cdot 2^{\frac{1}{2}}[/tex]

. . . . . . . . . [tex]n \;>\; 16\sqrt{2} \;=\; 22.627417[/tex]

Therefore: . [tex]n \;=\;23[/tex]

 
soroban said:
Hello, anemone!


kaliprasad is correct.
I used a different approach.

We want: .[tex]n^{16} \;> \; 16^{18}[/tex]

. . . . . . . .[tex]n^{16} \;>\; (2^4)^{18}[/tex]

. . . . . . . .[tex]n^{16} \;>\;2^{72}[/tex]

. . . . . . . . . [tex]n \;>\;2^{\frac{72}{16}}\;=\;2^{\frac{9}{2}}[/tex]

. . . . . . . . . [tex]n \;>\; 2^{4+\frac{1}{2}} \;=\;2^4 \cdot 2^{\frac{1}{2}}[/tex]

. . . . . . . . . [tex]n \;>\; 16\sqrt{2} \;=\; 22.627417[/tex]

Therefore: . [tex]n \;=\;23[/tex]

above approach is more straight forward