Question

Logarithm Question...

Given \(n^3 < 10^n\) for a fixed positive integer \( n \ge 1\) , prove that \((n+1)^3 < 10^{n+1}\)

Answer 1

Regarding this question, I tried :


\( \log n^3 < n \)

\(\log n^3 +1 < n + 1\)

\((\log n^3 + 1)^3 < (n+1)^3\)

\((\log(10n^3) )^3 < (n+1)^3\)

Not sure, whether it is the best way to do it, but, I think, we can get something from it.

Answer 2

induction hehe

Answer 3

And, if I just don't add 1 both sides and follow other method :

\(\log (n^3) < n \\

3\log (n) < n \\

3 < \cfrac{n}{\log n}\)

Do I get something from here? Well, i'm not sure.

Answer 4

why dnt u use induction :O

Answer 5

or suppose to use log*

Answer 6

Well, I would love to see how to do it with induction, but as log is more important for me (currently) , so I'm aiming it to solve by using Logarithm Properties.

Answer 7

I got something : \(\large n < 10^{n + \cfrac{1}{3}}\)

Answer 8

I see n^3 < 10^n for any n. :|
Proving this much should be sufficient.

Answer 9

For any n? Hmm... nope :/

Answer 10

put, m = n+1

Answer 11

Now, for any \(n \ge 1\), \(m \ge 2\).

Answer 12

Now,

\((m^3 < 10^m)\)

Since, m satisfies the condition \(m \ge 1\)

Answer 13

Now, again put m = n+1

Answer 14

You will get your required proof.

Answer 15

Oh :P .... I thought there will be a use of Logarithm... !

Answer 16

for me i would try this :-
for n=1 its true
lets it be true for n
show for n+1
(n+1)^3=n^3+3n^2+3n+1 =\( n^3(1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3})\)
the induction step
n^3<10^n eq 1
1<\(1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}\le\) 8 <10 eq 2

so
(n+1)3<10^n+1

Answer 17

No problem

Answer 18

@vishweshshrimali5 But that is still induction...

Answer 19

You can take the log of both sides and then choose \(m = n + 1\), but it makes on difference. So I guess your brother's technique is fine. :P

Answer 20

Tell you what, leave that method. Its wrong.

Answer 21

It never says that n is a general value. It can be some particular n.

Answer 22

Use only log there.

Answer 23

Yeah I knew it already :P xD

Answer 24

Yeah I just noticed that as well @vishweshshrimali5 because if it's for any integer n>1, then n+1 is technically already given to you and there's nothing to prove lol.

Answer 25

Yeah @Kainui, that's why it sounded fishy to me. ;)

Answer 26

Glad to know that the question is not that easy :P

@vishweshshrimali5 , @Kainui - the methods I used above, would they help us in leading to the answer? or they were just waste :P

Answer 27

I can't tell you until I figure it out, seems helpful though lol

Answer 28

Yep, sure, take your time.

Answer 29

Ok I think I got it.

Answer 30

From the first condition I get:

\[3 \times \log{n} < n\]
\[\implies 3\times \log{n} - n < 0\] ------------ (1)

Answer 31

Now, see that :

\[( 1+ \cfrac{1}{n}) > 1\]

As well as
\[ 2 \ge (1 + \cfrac{1}{n})\]

So,
\[8 \ge (1 + \cfrac{1}{n})^3 > 1\]

Answer 32

Now, thus,

\[(1 + \cfrac{1}{n})^3 < 10\]
\[\implies 3 \times \log{(1+\cfrac{1}{n})} < 1\]
\[\implies 3 \times \log{(\cfrac{n+1}{n})} < 1\]
\[\implies 3 \times \log{n+1} - 3 \times \log{n} < 1\]

Answer 33

Let this be equation (2)

Answer 34

Adding equation 1 and 2

Answer 35

I get,

\[3 \times \log{(n+1)} - n < 1\]
\[\implies 3 \times \log{(n+1)} < n+1\]
\[\implies (n+1)^3 < 10^{n+1}\]

HENCE PROVED

HURRAY !!!!

Answer 36

From where did you get this :
\(\left(1+\cfrac{1}{n}\right) ^3 < 10\) ?

Answer 37

Well, if \(2\ge(1+\cfrac{1}{n})\) then clearly cubing both sides I will get,

\[8 \ge (1+\cfrac{1}{n})^3\]

Now, if a number is equal to or less than 8, then it must be less than 10 so, I get the required expression.

Answer 38

The condition \(n \ge 1\) was very important if I wanted to make the argument that

\[(1+\cfrac{1}{n}) > 1\]

Answer 39

Got that ?

Answer 40

o.O yeah, I thought the signs were reversed :P

Answer 41

Nope

Answer 42

Yep.. got it now. Thanks a lot @vishweshshrimali5 - Awesome work...! :-)

Answer 43

A simple but nice question. Hats off to the one who made that question.

It was my pleasure @mathslover

Answer 44

Well actually, I am not wearing a hat right now. :P So, forgot that one.

Answer 45

*forget

Answer 46

:) Thanks everyone.