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Simplification of different base logarithms

Different log basesDoubt on Logarithms multiplicationYet another logarithm multiplication doubtWhat's a good class of functions for bounding/comparing ratios of complicated logarithms?Is this sum of 2 different irrational logarithms irrational: $log_2(3)+log_3(2)$?Simplification of a logarithm expressionIn Need of Logarithms Simplification ExercisesIntroduction to Logarithms and Natural LogsEvaluating $frac{1}{log_2 (100!)} + cdots + frac{1}{log_{100}(100!)}$Which of the following is bigger (logarithms)

3

3

$begingroup$

I'm in doubt on simplifying the expression:
$log_2 6 - log_4 9$

Working on it I've got: $log_2 6 - dfrac{log_2 9}{2}$

There's anyway to simplify it more ? I'm learning logarithms now so I'm not aware of all properties and tricks.

Thanks in advance

logarithms

share|cite|improve this question

edited Nov 18 '12 at 20:58

aajjbb

asked Nov 18 '12 at 20:35

aajjbbaajjbb

5252715

$endgroup$

add a comment | 

3

3

$begingroup$

I'm in doubt on simplifying the expression:
$log_2 6 - log_4 9$

Working on it I've got: $log_2 6 - dfrac{log_2 9}{2}$

There's anyway to simplify it more ? I'm learning logarithms now so I'm not aware of all properties and tricks.

Thanks in advance

logarithms

share|cite|improve this question

edited Nov 18 '12 at 20:58

aajjbb

asked Nov 18 '12 at 20:35

aajjbbaajjbb

5252715

$endgroup$

add a comment | 

$begingroup$

I'm in doubt on simplifying the expression:
$log_2 6 - log_4 9$

Working on it I've got: $log_2 6 - dfrac{log_2 9}{2}$

There's anyway to simplify it more ? I'm learning logarithms now so I'm not aware of all properties and tricks.

Thanks in advance

logarithms

share|cite|improve this question

edited Nov 18 '12 at 20:58

aajjbb

asked Nov 18 '12 at 20:35

aajjbbaajjbb

5252715

$endgroup$

share|cite|improve this question

edited Nov 18 '12 at 20:58

aajjbb

asked Nov 18 '12 at 20:35

aajjbbaajjbb

5252715

share|cite|improve this question

edited Nov 18 '12 at 20:58

aajjbb

asked Nov 18 '12 at 20:35

aajjbbaajjbb

5252715

asked Nov 18 '12 at 20:35

aajjbbaajjbb

5252715

asked Nov 18 '12 at 20:35

aajjbbaajjbb

5252715

3 Answers
3

active

oldest

votes

4

$begingroup$

You can do a bit more simplification. The important properties of the log function are, for any base $a>0$,

• $log_a(bc)=log_ab+log_ac$, so, for example, $log_26=log_22+log_23$


• $log_a(b/c)=log_ab-log_ac$


• $log_ab^n=nlog_ab$


• $log_ab=1/log_ba$


• $(log_ab)(log_bc)=log_ac$


• $log_aa=1$

So, for example, we can simplify $log_26-log_49$ as
$$
begin{align}
log_26-log_49&=log_2(2cdot3)-log_4(3^2) \
&= log_22+log_23-2log_43 &text{using the first and third identities}\
&=1+log_23-2log_43 &text{using the sixth identity}\
&=1+log_23-2log_42log_23 &text{using the fifth}\
&=1+log_23-2(log_23)/log_24 &text{using the fourth}\
&=1+log_23-2(log_23)/2 &text{using the third}\
&=1+log_23-log_23 &text{using a bit of algebra}\
&=1
end{align}
$$

share|cite|improve this answer

answered Nov 18 '12 at 21:03

Rick DeckerRick Decker

7,63432341

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, this table will helps me a lot. but in the Latex representation, the base is being shadowed by the 'G' of log.
  $endgroup$
  – aajjbb
  Nov 18 '12 at 21:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @aajbb Yeah, that happens to a lot of people, myself included. The consensus is that this site's LaTeX processor doesn't play nicely with some browsers. Often, if you simply tell the browser to refresh the page, the problem with overlapping text goes away.
  $endgroup$
  – Rick Decker
  Nov 18 '12 at 21:16
  

  
  
  
  

add a comment | 

2

$begingroup$

You're correct so far. Bring out the factor of $frac{1}{2}$ to get $frac{1}{2}(2log_{2}(6)-log_{2}(9))$. Since $alog(b)=log(b^{a})$ $log(a)-log(b)=log(frac{a}{b})$, you get $2log_{2}(6)=log_{2}(6^{2})$, and $$frac{1}{2}(log_{2}(36)-log_{2}(9))=frac{1}{2}left(log_{2}left(frac{36}{9}right)right)=log_{2}(4)/2=2/2=1 $$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

preferred_anonpreferred_anon

13k11743

$endgroup$

add a comment | 

1

$begingroup$

$log_26-frac{log_29}{2}=log_22+log_23-1/2×（2×log_23）=1$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

Ziqian XieZiqian Xie

604515

$endgroup$

add a comment | 

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4

$begingroup$

You can do a bit more simplification. The important properties of the log function are, for any base $a>0$,

• $log_a(bc)=log_ab+log_ac$, so, for example, $log_26=log_22+log_23$


• $log_a(b/c)=log_ab-log_ac$


• $log_ab^n=nlog_ab$


• $log_ab=1/log_ba$


• $(log_ab)(log_bc)=log_ac$


• $log_aa=1$

So, for example, we can simplify $log_26-log_49$ as
$$
begin{align}
log_26-log_49&=log_2(2cdot3)-log_4(3^2) \
&= log_22+log_23-2log_43 &text{using the first and third identities}\
&=1+log_23-2log_43 &text{using the sixth identity}\
&=1+log_23-2log_42log_23 &text{using the fifth}\
&=1+log_23-2(log_23)/log_24 &text{using the fourth}\
&=1+log_23-2(log_23)/2 &text{using the third}\
&=1+log_23-log_23 &text{using a bit of algebra}\
&=1
end{align}
$$

share|cite|improve this answer

answered Nov 18 '12 at 21:03

Rick DeckerRick Decker

7,63432341

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, this table will helps me a lot. but in the Latex representation, the base is being shadowed by the 'G' of log.
  $endgroup$
  – aajjbb
  Nov 18 '12 at 21:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @aajbb Yeah, that happens to a lot of people, myself included. The consensus is that this site's LaTeX processor doesn't play nicely with some browsers. Often, if you simply tell the browser to refresh the page, the problem with overlapping text goes away.
  $endgroup$
  – Rick Decker
  Nov 18 '12 at 21:16
  

  
  
  
  

add a comment | 

4

$begingroup$

You can do a bit more simplification. The important properties of the log function are, for any base $a>0$,

• $log_a(bc)=log_ab+log_ac$, so, for example, $log_26=log_22+log_23$


• $log_a(b/c)=log_ab-log_ac$


• $log_ab^n=nlog_ab$


• $log_ab=1/log_ba$


• $(log_ab)(log_bc)=log_ac$


• $log_aa=1$

So, for example, we can simplify $log_26-log_49$ as
$$
begin{align}
log_26-log_49&=log_2(2cdot3)-log_4(3^2) \
&= log_22+log_23-2log_43 &text{using the first and third identities}\
&=1+log_23-2log_43 &text{using the sixth identity}\
&=1+log_23-2log_42log_23 &text{using the fifth}\
&=1+log_23-2(log_23)/log_24 &text{using the fourth}\
&=1+log_23-2(log_23)/2 &text{using the third}\
&=1+log_23-log_23 &text{using a bit of algebra}\
&=1
end{align}
$$

share|cite|improve this answer

answered Nov 18 '12 at 21:03

Rick DeckerRick Decker

7,63432341

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, this table will helps me a lot. but in the Latex representation, the base is being shadowed by the 'G' of log.
  $endgroup$
  – aajjbb
  Nov 18 '12 at 21:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @aajbb Yeah, that happens to a lot of people, myself included. The consensus is that this site's LaTeX processor doesn't play nicely with some browsers. Often, if you simply tell the browser to refresh the page, the problem with overlapping text goes away.
  $endgroup$
  – Rick Decker
  Nov 18 '12 at 21:16
  

  
  
  
  

add a comment | 

$begingroup$

You can do a bit more simplification. The important properties of the log function are, for any base $a>0$,

• $log_a(bc)=log_ab+log_ac$, so, for example, $log_26=log_22+log_23$


• $log_a(b/c)=log_ab-log_ac$


• $log_ab^n=nlog_ab$


• $log_ab=1/log_ba$


• $(log_ab)(log_bc)=log_ac$


• $log_aa=1$

So, for example, we can simplify $log_26-log_49$ as
$$
begin{align}
log_26-log_49&=log_2(2cdot3)-log_4(3^2) \
&= log_22+log_23-2log_43 &text{using the first and third identities}\
&=1+log_23-2log_43 &text{using the sixth identity}\
&=1+log_23-2log_42log_23 &text{using the fifth}\
&=1+log_23-2(log_23)/log_24 &text{using the fourth}\
&=1+log_23-2(log_23)/2 &text{using the third}\
&=1+log_23-log_23 &text{using a bit of algebra}\
&=1
end{align}
$$

share|cite|improve this answer

answered Nov 18 '12 at 21:03

Rick DeckerRick Decker

7,63432341

$endgroup$

share|cite|improve this answer

answered Nov 18 '12 at 21:03

Rick DeckerRick Decker

7,63432341

answered Nov 18 '12 at 21:03

Rick DeckerRick Decker

7,63432341

answered Nov 18 '12 at 21:03

Rick DeckerRick Decker

7,63432341

2

$begingroup$

You're correct so far. Bring out the factor of $frac{1}{2}$ to get $frac{1}{2}(2log_{2}(6)-log_{2}(9))$. Since $alog(b)=log(b^{a})$ $log(a)-log(b)=log(frac{a}{b})$, you get $2log_{2}(6)=log_{2}(6^{2})$, and $$frac{1}{2}(log_{2}(36)-log_{2}(9))=frac{1}{2}left(log_{2}left(frac{36}{9}right)right)=log_{2}(4)/2=2/2=1 $$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

preferred_anonpreferred_anon

13k11743

$endgroup$

add a comment | 

2

$begingroup$

You're correct so far. Bring out the factor of $frac{1}{2}$ to get $frac{1}{2}(2log_{2}(6)-log_{2}(9))$. Since $alog(b)=log(b^{a})$ $log(a)-log(b)=log(frac{a}{b})$, you get $2log_{2}(6)=log_{2}(6^{2})$, and $$frac{1}{2}(log_{2}(36)-log_{2}(9))=frac{1}{2}left(log_{2}left(frac{36}{9}right)right)=log_{2}(4)/2=2/2=1 $$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

preferred_anonpreferred_anon

13k11743

$endgroup$

add a comment | 

$begingroup$

You're correct so far. Bring out the factor of $frac{1}{2}$ to get $frac{1}{2}(2log_{2}(6)-log_{2}(9))$. Since $alog(b)=log(b^{a})$ $log(a)-log(b)=log(frac{a}{b})$, you get $2log_{2}(6)=log_{2}(6^{2})$, and $$frac{1}{2}(log_{2}(36)-log_{2}(9))=frac{1}{2}left(log_{2}left(frac{36}{9}right)right)=log_{2}(4)/2=2/2=1 $$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

preferred_anonpreferred_anon

13k11743

$endgroup$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

preferred_anonpreferred_anon

13k11743

answered Nov 18 '12 at 20:44

preferred_anonpreferred_anon

13k11743

answered Nov 18 '12 at 20:44

preferred_anonpreferred_anon

13k11743

1

$begingroup$

$log_26-frac{log_29}{2}=log_22+log_23-1/2×（2×log_23）=1$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

Ziqian XieZiqian Xie

604515

$endgroup$

add a comment | 

1

$begingroup$

$log_26-frac{log_29}{2}=log_22+log_23-1/2×（2×log_23）=1$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

Ziqian XieZiqian Xie

604515

$endgroup$

add a comment | 

$begingroup$

$log_26-frac{log_29}{2}=log_22+log_23-1/2×（2×log_23）=1$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

Ziqian XieZiqian Xie

604515

$endgroup$

share|cite|improve this answer

answered Nov 18 '12 at 20:44

Ziqian XieZiqian Xie

604515

answered Nov 18 '12 at 20:44

Ziqian XieZiqian Xie

604515

answered Nov 18 '12 at 20:44

Ziqian XieZiqian Xie

604515