Is it very small!

Algebra Level 2

Find the largest integer $n$ for which $\displaystyle n^{200} < 5^{300}$ .

5 solutions

Naitik Sanghavi
Jul 24, 2015

n^200<5^300 (n²)^100<(5³)^100 Therefore n²<5³ n²<125

Therefore largest integer n =11

Try to learn some LaTeX...

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\ ( n^{200} < 5^{300} \ (n^2)^{100} < (5^3)^{100} \ n^2 < 5^3 \ n^2 < 125 \ max(n) = 125 \ )

And remove the spacing between "\ ( " and " \ )"

Pi Han Goh - 5 years, 10 months ago

Same way! I did it

Department 8 - 5 years, 10 months ago

The Adammz
Aug 8, 2015

${ n }^{ 200 }<{ 5 }^{ 300 }$

$200<\log _{ n }{ { 5 }^{ 300 } }$

$200<300 log _{ n }{ { 5 }}$

$\frac { 2 }{ 3 } <log _{ n }{ { 5 }}$

${ n }^{ \frac { 2 }{ 3 } } <{ { 5 } }$

${ n }<\sqrt { { 5 }^{ 3 } }$

${ n }<\sqrt { 125 }$

${ 11 }^{2 } <\sqrt { 125 } <{ 12 }^{ 2 }$

$\therefore n = 11$

I used the same idea, but with log5

$log_{5} ~ n ^{200} < log_{5}~ 5^{300}$

$200 * log_5 ~ n < 300 * log_5~ 5$

$log_5 ~ n < \frac{3}{2}$

$n < 5^ \frac{3}{2}$

Eduardo Augusto Sobral Junior - 5 years, 10 months ago

Ben Habeahan
Aug 18, 2015

$n^{200} < 5^{300} \\ (n^2)^{100} < (5^3)^{100} \\ n^2 < 5^3 \\ n^2 < 125 \\ n < \sqrt{125} \\ 11<\sqrt{125} <12 \\ so \boxed{n=11}$

Gokul Kumar
Jul 26, 2015

after taking log on both sides we get,

200 log n < 300 log 5 (log base 10)

2 log n <3 log 5

log n< 1.5 * 0.69897000433

log n < 1.0484550065

log 11 is smaller than that value so----> 11

Edwin Gray
Sep 15, 2018

For equality, 200 lof(n) = 300 log(5) , or n = 11.18. Therefore n = 11. Ed Gray