Question

A smarter way to do this logarithm problem

Answer 1

Prove that :-
\[\huge \log_710 > \log_{11}13\]

Answer 2

I was able to prove this but it was very lengthy method

Answer 3

For it to be true

\(\log_7 {10} - \log_{11} {13} > 0\)

Answer 4

yes

Answer 5

change the base then subtract using the homomorphism property , then solve:)

Answer 6

I think we have not learnt the homomorphism property , or maybe its a different name to what properties we have learnt, could you please explain what this property is about if possible

Answer 7

He just means \(\log(a ) - \log(b) = \log(a/b)\)... I guess...

Answer 8

Or,
log(a) + log(b) = log(ab) is that what you mean @zzr0ck3r

Answer 9

\(\cfrac{1}{\log_{10} 7} - \cfrac{\log_{10} 13 }{\log_{10} 11} > 0\)

\(\cfrac{\log_{10} 11 - \log_{10} 13 \times \log_{10} 7 }{\log_{10} 7 \times \log_{10} 11 } > 0\)

Answer 10

LOGIC :

Answer 11

\(\log_{10} 7 < 1\)

That is it will be a decimal with integral part = 0 .

Answer 12

yes @No.name this is the homomorphism property

Answer 13

\(2>\log _{10} 13 > 1\)

Answer 14

You can shift the denomiantor it would be always positive
And yes that's what i did but i took base as 7

Answer 15

@mathslover ^

Answer 16

Thus, by pure mathematical logic : (lol)

\(\log_{10} 13 \times \log_{10} 7 \) will be \(< 1\)

Answer 17

Greater than 0ne i suppose

Answer 18

But, \(\log_{10} 11\) > 1

Thus HENCE PROVED! :P

Answer 19

oh .. wait

Answer 20

You can use the "pure mathematical logic" this way too:\[\rm Prove~ ~ \log(11)>\log(7)\log(13)\]

Answer 21

I took the base as 7 , that was the whole mess , but i managed to prove it somehow

Answer 22

what happened @mathslover

Answer 23

I'm trying. Wait.

Answer 24

okay

Answer 25

if ur method goes more than 1.5 pages long then that's what i did

Answer 26

By "Pure Mathematical Logic"

Answer 27

Naah... its of 5-6 lines...

Answer 28

Clearly...

13 > 11
\(13^{\log_{10} 7 } < 11\)

Answer 29

Take log both sides

\(\log 7 \log 13 < log 11\)

Answer 30

Thus, proved.

Answer 31

Well, how did you prove that?

Answer 32

Hmm wait, lemme right the LaTeX thing.

Answer 33

* write

Answer 34

How would it be proved like that??

Answer 35

Ah! I'm trying my level best to prove that. :/

Answer 36

\[13 > 11\]\[\log_{\log_{10}7}13 < \log_{\log_{10}7}11\]\[\log13 < \dfrac{\log11 }{\log 7}\]=)

Answer 37

Welp.

Answer 38

Yup, things cancel out and we're left with that.

Answer 39

lol are we allowed to use calculator?

Answer 40

I guess now we've proved \(\log(13) \log(7) < \log(11)\), we are done with the question.

Answer 41

No lol , it came in IIT , so no calculator

Answer 42

well.. :P that was the smartest way to do it... use a calculator ... :P haha

Answer 43

haha

Answer 44

Its obvious that \(\log7\) will lie between 0 and 1.

\(\log {11} > 1\)

and so, \(\log_{11} 13 \) will lie between 1.1 - 1.3

Answer 45

yes.

Answer 46

whereas , \(\cfrac{1}{\log 7}\) > 1

Answer 47

I'm yet to think the best method for this.

Will post here when I get it.

@No.name

Answer 48

Okay. :)

Answer 49

See this :- some good properties you'll may not know
[This is not related to the question]
http://en.wikipedia.org/wiki/Logarithmic_identities