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nice man! i didnt think of this and tried logs! :D

Ajmal Siddiqui
- 7 years, 5 months ago

If a < b, than how can we say 5a > b .. Above, a = 5^12 and b = 2^28

Vineet Singla
- 7 years, 5 months ago

This was my original intent :)

Matthew Fan
- 7 years, 5 months ago

I simply used a calculator, put in all the options successively :)

Aun Abbas Jaffery
- 7 years, 4 months ago

Just because $5^{12}<2^{28}$ doesn't mean that $5^{13}>2^{28}$ . To complete your proof, state that $5^4>2^7$ and multiply the inequalities $5^4\cdot5^3\cdot5^3\cdot5^3=5^{13}>2^7\cdot2^7\cdot2^7\cdot2^7=2^{28}$ .

Cody Johnson
- 7 years, 3 months ago

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Why would the $5^{4} > 2^{7}$ statement be enough? It doesn't seem it covers it fully. I'm not sure there's a proof that goes directly to 13 as a sum to the 28. One way to do this would be to note that $5^{14} > 2^{28}$ since $2^{28} = (2^{2})^{14} = 4^{14}$ and $5 > 4$ . Then you're left with either 13 or 14, and you can calculate to test it.

Felipe Perestrelo
- 5 years, 10 months ago

A very nice solution

Devesh Rai
- 7 years, 5 months ago

nice

Tirtham Mukherjee
- 7 years, 5 months ago

what a romantic solution! :))

Brandon Chow
- 7 years, 5 months ago

i was near just stuck at 5th step :(

Saloni Baweja
- 7 years, 5 months ago

i too tried logs

Divya Sitani
- 7 years, 5 months ago

good one...

Amlan Mishra
- 7 years, 5 months ago

I have tried for 5^1>2^2.. Damn!!!

Saikat Chakrabortty
- 7 years, 4 months ago

really very nice...the only catch in this reasoning is that if a^x>b and a^y>b, then equating x and y would fail if we deal with rational numbers.

kalyan pakala
- 7 years, 4 months ago

I never thought that this problem could be solved so easily.

Partho Kunda
- 7 years, 4 months ago

It's great.

A Former Brilliant Member
- 7 years, 4 months ago

nice one!!!!!!!!

Suyash Mishra
- 7 years, 2 months ago

I wrote a problem almost exactly like this, except it actually used logarithms. Nice solution, though. :)

Asher Joy
- 7 years ago

Jaja damm.... I did the same but imao I did It wrong at the end haha I though it was 42 haha I got it bad because I didn't pay attention.

Axel Ordóñez
- 5 years, 10 months ago

i try logs, but i dont know how much log0.4 is.So.........

Wang Yi
- 5 years, 10 months ago

We simply apply the logarithmic method.

so,

5^n > 10^28

(log both the sides)

= log (5^n) > log (10^28)

= n (log 5) > 28

= n > 28 / (log 5)

= n > 40.0589...

Therefore, n = 41 is the minimum

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Good.

Soham Dibyachintan
- 7 years, 5 months ago

You need to note the monotonicity of the logarithm function to verify the transformation of the inequality.

Cody Johnson
- 7 years, 3 months ago

great lol

Kishore Kumar
- 7 years, 5 months ago

Taking log 10 of both sides, we have $n \log 5>28 \log 10=28 \Rightarrow n>\frac{28}{\log 5}$ . Since $\log 5$ is just nearly under $\frac{7}{10}$ , we have $n>40.some$ , so the least integer that satisfies this is $\lceil 40.some \rceil$ , which is creary (hopefully?) $\boxed{41}$

(Note that an exacter (?) value of $\log 5$ can be found using a calculator, and that it is a good idea to memorise that $\log 5$ is slightly under $\frac{7}{10}$ (if you haven't already))

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I found a way to improve this solution a bit..The title of the question solves the problem.

$5^n>10^{28}\implies 5^n>5^{28}\cdot 2^{28}$

This implies that we are looking for a $k$ such that

$5^k>2^{28}=(2^7)^4=(128)^4=(125+3)^4=(5^3+3)^4= (5^{12}+4*5^9*3+6*5^6*3^2+ 4*5^3*3^3+3^4)$

This considerably reduces the number of cases to check,as the left hand side must be greater than the first term of the expansion( $5^{12})$ .

Rahul Saha
- 7 years, 5 months ago

I did in the same way.. :)

Manish Mishra
- 7 years, 5 months ago

By taking log on both the sides, we get

n
*
log5 > 28
*
log10

n*log5 > 28

n > 28 / log5

n > 40.2

Hence n = 41

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5^{n} > 10^{28) Take log for both sides
*

*
nlog5 > 28log10 divide by log5 (log10 = 1)
*

*
n > 28/{log5} which is approximately equals 40
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*
then the least integer value of n = 41
*

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Well, obviously $5^{n} > 10^{28} \Rightarrow 5^{n-28} > 2^{28}$ . Now, if we apply the natural log function on both sides we get

$n = \left \lceil 28 + \frac{\ln2^{28}}{\ln5} \right \rceil = 41$

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$5^{n} > 10^{28}$$5^{n} > (5 \times 2)^{28}$$5^{n} > 5^{28} \times 2^{28}$$5^{n-28} > 2^{28}$Note that $5^{3} < 2^{7}$ .

Have both sides multiplied to the fourth power, then $5^{12} < 2^{28}$ .

Therefore,

$5^{13} > 2^{28}$Since

n - 28 = 13, then n = 41