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Solve it!

Sovr'gnChancellor£

Solve this:

Find x > 3 such that

ln(x) < x^(0.1)

Reply 1

PrinceyJ

Original post by Sovr'gnChancellor£

Solve this:

I assume you're looking for an integer solution?

Reply 2

Sovr'gnChancellor£

Original post by PrinceyJ

I assume you're looking for an integer solution?

Yes.

Reply 3

dragonrabbit

The mathematician in me should be ashamed :tongue:

But I've tried all integers up to 6829 and all the answers for x^(0.1)-ln(x) are negative, and rising (in magnitude) Are you sure there's an answer? :biggrin:

Edit: Googled it and found an answer on yahoo :s-smilie:

Now to find it myself (I didn't look at the working!)

(edited 11 years ago)

Reply 4

PrinceyJ

Original post by Sovr'gnChancellor£

Yes.

There aren't any below 21 Billion (and then my computer crashed):

#include <iostream>
#include <math.h>
usingnamespacestd;


int main(int argc, const char * argv[])
{
    long double x = 3;

    long double lr = 1, rr = 0;

    while (rr<lr){
        x+=1;
        lr = log(x);
        rr = pow(x, 0.1);
        if (lr<rr){
            cout << "Answer: " << x << "lr:" << lr << "rr:" << rr << endl;
        }else {
            cout << "Tried: " << x << "lr:" << lr << "rr:" << rr << endl;
        }
    }


    return 0;
}

Nor does it look like it's going to converge

Reply 5

dragonrabbit

Original post by PrinceyJ

It's really big!!! Really really big! I want to know how they worked it out :s-smilie:

But don't want to just look at their working :tongue:

Wish I understood these logs and e's :tongue:

The answer doesn't actually have a proper proof/range, I'm going to have a ponder... :s-smilie:

(edited 11 years ago)

Reply 6

danny111

Original post by Sovr'gnChancellor£

Solve this:

Have you tried looking at the derivatives?

Reply 7

PrinceyJ

Original post by dragonrabbit

It's really big!!! Really really big! I want to know how they worked it out :s-smilie:

But don't want to just look at their working :tongue:

Wish I understood these logs and e's :tongue:

Woah, it starts to converge!

Reply 8

Jtking3000

ln(x)-x^0.1=0
dy/dx=x^-1-0.1*x^-0.9=0

Looks like the number will have to be bigger than 10^10...

Reply 9

dragonrabbit

Q ^^ Can you just differentiate/integrate whenever you please (both sides) just as you would multiplying 2 sides of an equation etc.
I should probably already know this but hey ^^ :biggrin:

STEP will be fun :tongue:

Reply 10

Jtking3000

Original post by dragonrabbit

Q ^^ Can you just differentiate/integrate whenever you please (both sides) just as you would multiplying 2 sides of an equation etc.
I should probably already know this but hey ^^ :biggrin:

STEP will be fun :tongue:

I assume you can...although I'm an engineering student not a maths student so I might get scorn for treating an inequality as an normal equation :tongue:

But it still holds true that 10^10 is the tipping point.

Reply 11

IrrationalNumber

It's not as hard as it looks! It's far easier if you take exponentials, and then remember that

e^{x} \geq \frac{x^r}{r!}

for any r. If you choose

r

sufficiently large, you can kill the denominator of the fraction you get when you take exponentials, and then you need to take x sufficiently large as well. But those are the basic ideas.

Reply 12

IrrationalNumber

Actually, the question seems to serve two purposes
1) To show you that

\log(x)

grows slower than any polynomial with a positive leading coefficient
2) To show you that for mathematics problems, it does not suffice to just try the first 1000000 numbers and conclude that it holds all the time.

Reply 13

Sovr'gnChancellor£

Original post by IrrationalNumber

It's not as hard as it looks! It's far easier if you take exponentials, and then remember that

e^{x} \geq \frac{x^r}{r!}

for any r. If you choose

r

sufficiently large, you can kill the denominator of the fraction you get when you take exponentials, and then you need to take x sufficiently large as well. But those are the basic ideas.