Prove that $log_23>log_35>log_47$.

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Using calculus, prove that $log_23>log_35>log_47$ .

My try :

If $log_x(2x-1)$ is decreasing function then we can say that $log_23>log_35>log_47$.

$f(x)=log_x(2x-1)$

$f(x)=dfracln(2x-1)ln x$

$f'(x)=dfrac2xln x-(x-1)ln(2x-1)(ln x)^2x(2x-1)$

calculus logarithms

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edited Jul 27 at 1:31

Nosrati

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asked Jul 26 at 3:27

Koolman

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up vote
6
down vote

favorite

Using calculus, prove that $log_23>log_35>log_47$ .

My try :

If $log_x(2x-1)$ is decreasing function then we can say that $log_23>log_35>log_47$.

$f(x)=log_x(2x-1)$

$f(x)=dfracln(2x-1)ln x$

$f'(x)=dfrac2xln x-(x-1)ln(2x-1)(ln x)^2x(2x-1)$

calculus logarithms

share|cite|improve this question

edited Jul 27 at 1:31

Nosrati

19.2k41544

asked Jul 26 at 3:27

Koolman

1,472617

add a comment | 

up vote
6
down vote

favorite

up vote
6
down vote

favorite

Using calculus, prove that $log_23>log_35>log_47$ .

My try :

If $log_x(2x-1)$ is decreasing function then we can say that $log_23>log_35>log_47$.

$f(x)=log_x(2x-1)$

$f(x)=dfracln(2x-1)ln x$

$f'(x)=dfrac2xln x-(x-1)ln(2x-1)(ln x)^2x(2x-1)$

calculus logarithms

share|cite|improve this question

edited Jul 27 at 1:31

Nosrati

19.2k41544

asked Jul 26 at 3:27

Koolman

1,472617

Using calculus, prove that $log_23>log_35>log_47$ .

My try :

If $log_x(2x-1)$ is decreasing function then we can say that $log_23>log_35>log_47$.

$f(x)=log_x(2x-1)$

$f(x)=dfracln(2x-1)ln x$

$f'(x)=dfrac2xln x-(x-1)ln(2x-1)(ln x)^2x(2x-1)$

calculus logarithms

share|cite|improve this question

edited Jul 27 at 1:31

Nosrati

19.2k41544

asked Jul 26 at 3:27

Koolman

1,472617

share|cite|improve this question

share|cite|improve this question

edited Jul 27 at 1:31

Nosrati

19.2k41544

edited Jul 27 at 1:31

Nosrati

19.2k41544

edited Jul 27 at 1:31

Nosrati

19.2k41544

19.2k41544

asked Jul 26 at 3:27

Koolman

1,472617

asked Jul 26 at 3:27

Koolman

1,472617

asked Jul 26 at 3:27

Koolman

1,472617

1,472617

add a comment | 

add a comment | 

3 Answers
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4
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From your computation of $f'(x)$ (after fixing a small typo: $x-1$ should be $2x-1$), it remains to show $$2 x log x < (2x-1) log (2x-1).tag$*$$$

The derivative of $g(x) = (2x-1) log (2x-1) - 2 x log x$ is $$g'(x) = 2 log(2x-1) + 2 - 2 (log x + 1) = 2 log (2 - frac1x),$$
which is negative for $1/2 < x < 1$ and positive for $x > 1$. Thus $g$ is minimized at $x=1$ with value $0$, and thus ($*$) holds for all $x > 1/2$ except at $x=1$, where both sides are equal.

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answered Jul 26 at 4:16

angryavian

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You might need to do your derivative again. The numerator of the derivative of $fracln(2x-1)ln x$ obtained by the quotient rule is
$$frac12x-1ln x-frac2x-1x$$
which we can upper-bound as
$$frac12x-1ln x-frac2x-1xlefrac x2x-1-frac2x-1x=frac-3x^2+4x-1x(2x-1)$$
The denominator of this last fraction is positive for $x>frac12$ and the numerator is negative for $x>1$. Therefore the whole derivative must be negative for $2le xle 4$, the interval of interest to the question, and $log_x(2x-1)$ is decreasing over $(2,4)$, so $log_23>log_35>log_47$.

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answered Jul 26 at 3:44

Parcly Taxel

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up vote
1
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Again $g(x)=2xln x-(2x-1)ln(2x-1)$ is a decreasing function in $(1,infty)$ because $g'(x)<0$ then from $x>1$ we have
$$2xln x-(2x-1)ln(2x-1)<0$$
then $f'(x)>0$.

share|cite|improve this answer

edited Jul 26 at 4:10

answered Jul 26 at 3:45

Nosrati

19.2k41544

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

From your computation of $f'(x)$ (after fixing a small typo: $x-1$ should be $2x-1$), it remains to show $$2 x log x < (2x-1) log (2x-1).tag$*$$$

The derivative of $g(x) = (2x-1) log (2x-1) - 2 x log x$ is $$g'(x) = 2 log(2x-1) + 2 - 2 (log x + 1) = 2 log (2 - frac1x),$$
which is negative for $1/2 < x < 1$ and positive for $x > 1$. Thus $g$ is minimized at $x=1$ with value $0$, and thus ($*$) holds for all $x > 1/2$ except at $x=1$, where both sides are equal.

share|cite|improve this answer

answered Jul 26 at 4:16

angryavian

34.6k12874

add a comment | 

up vote
4
down vote

From your computation of $f'(x)$ (after fixing a small typo: $x-1$ should be $2x-1$), it remains to show $$2 x log x < (2x-1) log (2x-1).tag$*$$$

The derivative of $g(x) = (2x-1) log (2x-1) - 2 x log x$ is $$g'(x) = 2 log(2x-1) + 2 - 2 (log x + 1) = 2 log (2 - frac1x),$$
which is negative for $1/2 < x < 1$ and positive for $x > 1$. Thus $g$ is minimized at $x=1$ with value $0$, and thus ($*$) holds for all $x > 1/2$ except at $x=1$, where both sides are equal.

share|cite|improve this answer

answered Jul 26 at 4:16

angryavian

34.6k12874

add a comment | 

up vote
4
down vote

up vote
4
down vote

From your computation of $f'(x)$ (after fixing a small typo: $x-1$ should be $2x-1$), it remains to show $$2 x log x < (2x-1) log (2x-1).tag$*$$$

The derivative of $g(x) = (2x-1) log (2x-1) - 2 x log x$ is $$g'(x) = 2 log(2x-1) + 2 - 2 (log x + 1) = 2 log (2 - frac1x),$$
which is negative for $1/2 < x < 1$ and positive for $x > 1$. Thus $g$ is minimized at $x=1$ with value $0$, and thus ($*$) holds for all $x > 1/2$ except at $x=1$, where both sides are equal.

share|cite|improve this answer

answered Jul 26 at 4:16

angryavian

34.6k12874

From your computation of $f'(x)$ (after fixing a small typo: $x-1$ should be $2x-1$), it remains to show $$2 x log x < (2x-1) log (2x-1).tag$*$$$

The derivative of $g(x) = (2x-1) log (2x-1) - 2 x log x$ is $$g'(x) = 2 log(2x-1) + 2 - 2 (log x + 1) = 2 log (2 - frac1x),$$
which is negative for $1/2 < x < 1$ and positive for $x > 1$. Thus $g$ is minimized at $x=1$ with value $0$, and thus ($*$) holds for all $x > 1/2$ except at $x=1$, where both sides are equal.

share|cite|improve this answer

answered Jul 26 at 4:16

angryavian

34.6k12874

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share|cite|improve this answer

answered Jul 26 at 4:16

angryavian

34.6k12874

answered Jul 26 at 4:16

angryavian

34.6k12874

answered Jul 26 at 4:16

angryavian

34.6k12874

34.6k12874

add a comment | 

add a comment | 

up vote
1
down vote

You might need to do your derivative again. The numerator of the derivative of $fracln(2x-1)ln x$ obtained by the quotient rule is
$$frac12x-1ln x-frac2x-1x$$
which we can upper-bound as
$$frac12x-1ln x-frac2x-1xlefrac x2x-1-frac2x-1x=frac-3x^2+4x-1x(2x-1)$$
The denominator of this last fraction is positive for $x>frac12$ and the numerator is negative for $x>1$. Therefore the whole derivative must be negative for $2le xle 4$, the interval of interest to the question, and $log_x(2x-1)$ is decreasing over $(2,4)$, so $log_23>log_35>log_47$.

share|cite|improve this answer

answered Jul 26 at 3:44

Parcly Taxel

33.5k136588

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up vote
1
down vote

You might need to do your derivative again. The numerator of the derivative of $fracln(2x-1)ln x$ obtained by the quotient rule is
$$frac12x-1ln x-frac2x-1x$$
which we can upper-bound as
$$frac12x-1ln x-frac2x-1xlefrac x2x-1-frac2x-1x=frac-3x^2+4x-1x(2x-1)$$
The denominator of this last fraction is positive for $x>frac12$ and the numerator is negative for $x>1$. Therefore the whole derivative must be negative for $2le xle 4$, the interval of interest to the question, and $log_x(2x-1)$ is decreasing over $(2,4)$, so $log_23>log_35>log_47$.

share|cite|improve this answer

answered Jul 26 at 3:44

Parcly Taxel

33.5k136588

add a comment | 

up vote
1
down vote

up vote
1
down vote

You might need to do your derivative again. The numerator of the derivative of $fracln(2x-1)ln x$ obtained by the quotient rule is
$$frac12x-1ln x-frac2x-1x$$
which we can upper-bound as
$$frac12x-1ln x-frac2x-1xlefrac x2x-1-frac2x-1x=frac-3x^2+4x-1x(2x-1)$$
The denominator of this last fraction is positive for $x>frac12$ and the numerator is negative for $x>1$. Therefore the whole derivative must be negative for $2le xle 4$, the interval of interest to the question, and $log_x(2x-1)$ is decreasing over $(2,4)$, so $log_23>log_35>log_47$.

share|cite|improve this answer

answered Jul 26 at 3:44

Parcly Taxel

33.5k136588

You might need to do your derivative again. The numerator of the derivative of $fracln(2x-1)ln x$ obtained by the quotient rule is
$$frac12x-1ln x-frac2x-1x$$
which we can upper-bound as
$$frac12x-1ln x-frac2x-1xlefrac x2x-1-frac2x-1x=frac-3x^2+4x-1x(2x-1)$$
The denominator of this last fraction is positive for $x>frac12$ and the numerator is negative for $x>1$. Therefore the whole derivative must be negative for $2le xle 4$, the interval of interest to the question, and $log_x(2x-1)$ is decreasing over $(2,4)$, so $log_23>log_35>log_47$.

share|cite|improve this answer

answered Jul 26 at 3:44

Parcly Taxel

33.5k136588

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 3:44

Parcly Taxel

33.5k136588

answered Jul 26 at 3:44

Parcly Taxel

33.5k136588

answered Jul 26 at 3:44

Parcly Taxel

33.5k136588

33.5k136588

add a comment | 

add a comment | 

up vote
1
down vote

Again $g(x)=2xln x-(2x-1)ln(2x-1)$ is a decreasing function in $(1,infty)$ because $g'(x)<0$ then from $x>1$ we have
$$2xln x-(2x-1)ln(2x-1)<0$$
then $f'(x)>0$.

share|cite|improve this answer

edited Jul 26 at 4:10

answered Jul 26 at 3:45

Nosrati

19.2k41544

add a comment | 

up vote
1
down vote

Again $g(x)=2xln x-(2x-1)ln(2x-1)$ is a decreasing function in $(1,infty)$ because $g'(x)<0$ then from $x>1$ we have
$$2xln x-(2x-1)ln(2x-1)<0$$
then $f'(x)>0$.

share|cite|improve this answer

edited Jul 26 at 4:10

answered Jul 26 at 3:45

Nosrati

19.2k41544

add a comment | 

up vote
1
down vote

up vote
1
down vote

Again $g(x)=2xln x-(2x-1)ln(2x-1)$ is a decreasing function in $(1,infty)$ because $g'(x)<0$ then from $x>1$ we have
$$2xln x-(2x-1)ln(2x-1)<0$$
then $f'(x)>0$.

share|cite|improve this answer

edited Jul 26 at 4:10

answered Jul 26 at 3:45

Nosrati

19.2k41544

Again $g(x)=2xln x-(2x-1)ln(2x-1)$ is a decreasing function in $(1,infty)$ because $g'(x)<0$ then from $x>1$ we have
$$2xln x-(2x-1)ln(2x-1)<0$$
then $f'(x)>0$.

share|cite|improve this answer

edited Jul 26 at 4:10

answered Jul 26 at 3:45

Nosrati

19.2k41544

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 4:10

edited Jul 26 at 4:10

edited Jul 26 at 4:10

answered Jul 26 at 3:45

Nosrati

19.2k41544

answered Jul 26 at 3:45

Nosrati

19.2k41544

answered Jul 26 at 3:45

Nosrati

19.2k41544

19.2k41544

add a comment | 

add a comment | 

 

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