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4

$begingroup$

I ran into this issue during my homework. Using the rules of logarithms, I need to prove that
$$
-2lnbigg(frac{2}{sqrt{6}}bigg)=ln3-ln2
$$
So here were my steps:

1. First step:
   $$
   -2lnbigg(frac{2}{sqrt{6}}bigg)=lnbigg(frac{2}{sqrt{6}}bigg)^{-2}
   $$
   And that's as far as I got, because now I want to use the form $ln(a/b) = ln(a) - ln(b)$, but first I need to reduce the fraction because it is raised to the $-2$.

How do I evaluate $bigg(frac{2}{sqrt{6}}bigg)^{-2}$ ?

Thanks

logarithms

share|cite|improve this question

edited 9 hours ago

Daniele Tampieri

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asked 9 hours ago

Miguel AragonMiguel Aragon

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Miguel Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $x^{-a} = frac{1}{x^a}$
  $endgroup$
  – desiigner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited your title, as you had it backwards.
  $endgroup$
  – Acccumulation
  9 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

I ran into this issue during my homework. Using the rules of logarithms, I need to prove that
$$
-2lnbigg(frac{2}{sqrt{6}}bigg)=ln3-ln2
$$
So here were my steps:

1. First step:
   $$
   -2lnbigg(frac{2}{sqrt{6}}bigg)=lnbigg(frac{2}{sqrt{6}}bigg)^{-2}
   $$
   And that's as far as I got, because now I want to use the form $ln(a/b) = ln(a) - ln(b)$, but first I need to reduce the fraction because it is raised to the $-2$.

How do I evaluate $bigg(frac{2}{sqrt{6}}bigg)^{-2}$ ?

Thanks

logarithms

share|cite|improve this question

edited 9 hours ago

Daniele Tampieri

2,98244 gold badges1010 silver badges2323 bronze badges

asked 9 hours ago

Miguel AragonMiguel Aragon

2344 bronze badges

New contributor

Miguel Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that $x^{-a} = frac{1}{x^a}$
  $endgroup$
  – desiigner
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited your title, as you had it backwards.
  $endgroup$
  – Acccumulation
  9 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I ran into this issue during my homework. Using the rules of logarithms, I need to prove that
$$
-2lnbigg(frac{2}{sqrt{6}}bigg)=ln3-ln2
$$
So here were my steps:

1. First step:
   $$
   -2lnbigg(frac{2}{sqrt{6}}bigg)=lnbigg(frac{2}{sqrt{6}}bigg)^{-2}
   $$
   And that's as far as I got, because now I want to use the form $ln(a/b) = ln(a) - ln(b)$, but first I need to reduce the fraction because it is raised to the $-2$.

How do I evaluate $bigg(frac{2}{sqrt{6}}bigg)^{-2}$ ?

Thanks

logarithms

share|cite|improve this question

edited 9 hours ago

Daniele Tampieri

2,98244 gold badges1010 silver badges2323 bronze badges

asked 9 hours ago

Miguel AragonMiguel Aragon

2344 bronze badges

New contributor

Miguel Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 9 hours ago

Daniele Tampieri

2,98244 gold badges1010 silver badges2323 bronze badges

asked 9 hours ago

Miguel AragonMiguel Aragon

2344 bronze badges

New contributor

Miguel Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 9 hours ago

Daniele Tampieri

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asked 9 hours ago

Miguel AragonMiguel Aragon

2344 bronze badges

New contributor

Miguel Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 9 hours ago

Daniele Tampieri

2,98244 gold badges1010 silver badges2323 bronze badges

edited 9 hours ago

Daniele Tampieri

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asked 9 hours ago

Miguel AragonMiguel Aragon

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Miguel Aragon is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 9 hours ago

Miguel AragonMiguel Aragon

2344 bronze badges

5 Answers
5

active

oldest

votes

2

$begingroup$

$a^{-k} = frac 1{a^k}$ by definition.

So $bigg(frac{2}{sqrt{6}}bigg)^{-2} =frac 1{bigg(frac{2}{sqrt{6}}bigg)^{2}}=$

$frac 1{bigg(frac {2^2}{sqrt 6^2}bigg)}=frac {sqrt 6^2}{2^2}=frac 64=frac 32$

It will help to realize that $(frac ab)^{-1} = frac 1{(frac ab)} = frac ba$ and that $(frac ab)^k = frac {a^k}{b^k}$ to realize that that means $(frac ab)^{-k} = frac 1{(frac ab)^k}= frac 1{(frac {a^k}{b^k})} = frac {b^k}{a^k}$.

(Also $(frac ab)^{-k} = [(frac ab)^{-1}]^k = (frac ba)^k=frac {b^k}{a^k}$ or that $(frac ab)^{-k} = frac {a^{-k}}{b^{-k}} = frac {(frac 1{a^k})}{(frac 1{b^k})} = frac {b^k}{a^k}$.)

In any event

$(frac {2}{sqrt 6})^{-2} = (frac {sqrt 6}{ 2})^2 = frac {sqrt 6^2}{2^2} = frac 64 = frac 32$.

share|cite|improve this answer

answered 8 hours ago

fleabloodfleablood

75.5k22 gold badges2828 silver badges9494 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, this is the explanation I was looking for. So now after a few steps, I'll get ln(3/2) = ln(3) - ln(2).
  $endgroup$
  – Miguel Aragon
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

6

$begingroup$

$$begin{align}
-2ln left( frac{2}{sqrt 6} right) &= -2big( ln(2)-ln(sqrt{6}) big) \
&= -2ln(2)+2ln(6^{1/2}) \
&= -2ln(2)+ln(2cdot 3) \
&=-2ln(2)+ big( ln(2)+ln(3)big) \
&=ln(3)-ln(2)
end{align}$$
And for your specific question, remember that
$$left( frac{a}{b} right)^{-n}=left( frac{b}{a} right)^n$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Azif00Azif00

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$endgroup$

add a comment | 

1

$begingroup$

Using your first step,

$-2 ln(frac{2}{sqrt{6}}) = ln(frac{2}{sqrt{6}})^{-2} = ln frac{1}{(frac{2}{sqrt{6}})^{2}} = ln frac{6}{4} = lnfrac{3}{2} = ln 3 - ln 2$

share|cite|improve this answer

answered 8 hours ago

Jaime Grimal AlvesJaime Grimal Alves

23877 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Well you may start by distributing the index since $2$ and $sqrt 6$ are positive. Thus $$left(frac{2}{sqrt{6}}right)^{-2}=frac{2^{-2}}{{sqrt 6}^{-2}}.$$

Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^{-n}=frac {1}{a^n}.$$ Applying this to your expression, we have $$frac{2^{-2}}{{sqrt 6}^{-2}}=frac{frac {1}{2^2}}{frac{1}{{sqrt 6}^2}}=frac{frac {1}{4}}{frac{1}{6}}=frac{6}{4}=frac 32.$$

share|cite|improve this answer

answered 8 hours ago

AllawonderAllawonder

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$endgroup$

add a comment | 

0

$begingroup$

$$ -2 ln left( frac{2}{sqrt{6}} right) = ln 3 - ln 2$$

if and only if

$$ ln left( frac{2}{sqrt{6}} right)^{-2} = ln left( frac{3}{2}right) $$

if and only if

$$ ln left[ frac{1}{left(frac{2}{sqrt{6}}right)^{2}} right] = ln left( frac{3}{2}right)$$

And so, we have

$$ ln left( frac{6}{4} right) = ln left( frac{3}{2}right)$$

which is true.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

mlchristiansmlchristians

99422 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse
  $endgroup$
  – peek-a-boo
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to include them initially.
  $endgroup$
  – mlchristians
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $4 = 10 implies 7-3 =7+3 implies-3= 3 implies (-3)^2 = 3^2 implies 9=9$ which is true. Therefore $4 = 10$. Or if penguins were elephants then they would both eat food. Which is true. So penguins are elephants.
  $endgroup$
  – fleablood
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained." But if you do that you MUST specify the steps are actually if and only if steps. If you do not SPECIFICALLY state that it is reasonable to assume they are only forward implications and the proof is not valid.
  $endgroup$
  – fleablood
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your first comment is not if and only if; and neither is you second. Forward implications is one-way; necessary and sufficient (i.e., if and only if are both ways). In the case of this problem, start from the given equation and work it to where I finished. Any problem with that. Now, if you want to do the work, start w/ the last statement and by a series of implications, show it implies the initial equation. I chose iff and only if instead.
  $endgroup$
  – mlchristians
  8 hours ago
  

  
  
  
  

add a comment | 

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2

$begingroup$

$a^{-k} = frac 1{a^k}$ by definition.

So $bigg(frac{2}{sqrt{6}}bigg)^{-2} =frac 1{bigg(frac{2}{sqrt{6}}bigg)^{2}}=$

$frac 1{bigg(frac {2^2}{sqrt 6^2}bigg)}=frac {sqrt 6^2}{2^2}=frac 64=frac 32$

It will help to realize that $(frac ab)^{-1} = frac 1{(frac ab)} = frac ba$ and that $(frac ab)^k = frac {a^k}{b^k}$ to realize that that means $(frac ab)^{-k} = frac 1{(frac ab)^k}= frac 1{(frac {a^k}{b^k})} = frac {b^k}{a^k}$.

(Also $(frac ab)^{-k} = [(frac ab)^{-1}]^k = (frac ba)^k=frac {b^k}{a^k}$ or that $(frac ab)^{-k} = frac {a^{-k}}{b^{-k}} = frac {(frac 1{a^k})}{(frac 1{b^k})} = frac {b^k}{a^k}$.)

In any event

$(frac {2}{sqrt 6})^{-2} = (frac {sqrt 6}{ 2})^2 = frac {sqrt 6^2}{2^2} = frac 64 = frac 32$.

share|cite|improve this answer

answered 8 hours ago

fleabloodfleablood

75.5k22 gold badges2828 silver badges9494 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, this is the explanation I was looking for. So now after a few steps, I'll get ln(3/2) = ln(3) - ln(2).
  $endgroup$
  – Miguel Aragon
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$a^{-k} = frac 1{a^k}$ by definition.

So $bigg(frac{2}{sqrt{6}}bigg)^{-2} =frac 1{bigg(frac{2}{sqrt{6}}bigg)^{2}}=$

$frac 1{bigg(frac {2^2}{sqrt 6^2}bigg)}=frac {sqrt 6^2}{2^2}=frac 64=frac 32$

It will help to realize that $(frac ab)^{-1} = frac 1{(frac ab)} = frac ba$ and that $(frac ab)^k = frac {a^k}{b^k}$ to realize that that means $(frac ab)^{-k} = frac 1{(frac ab)^k}= frac 1{(frac {a^k}{b^k})} = frac {b^k}{a^k}$.

(Also $(frac ab)^{-k} = [(frac ab)^{-1}]^k = (frac ba)^k=frac {b^k}{a^k}$ or that $(frac ab)^{-k} = frac {a^{-k}}{b^{-k}} = frac {(frac 1{a^k})}{(frac 1{b^k})} = frac {b^k}{a^k}$.)

In any event

$(frac {2}{sqrt 6})^{-2} = (frac {sqrt 6}{ 2})^2 = frac {sqrt 6^2}{2^2} = frac 64 = frac 32$.

share|cite|improve this answer

answered 8 hours ago

fleabloodfleablood

75.5k22 gold badges2828 silver badges9494 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perfect, this is the explanation I was looking for. So now after a few steps, I'll get ln(3/2) = ln(3) - ln(2).
  $endgroup$
  – Miguel Aragon
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

$a^{-k} = frac 1{a^k}$ by definition.

So $bigg(frac{2}{sqrt{6}}bigg)^{-2} =frac 1{bigg(frac{2}{sqrt{6}}bigg)^{2}}=$

$frac 1{bigg(frac {2^2}{sqrt 6^2}bigg)}=frac {sqrt 6^2}{2^2}=frac 64=frac 32$

It will help to realize that $(frac ab)^{-1} = frac 1{(frac ab)} = frac ba$ and that $(frac ab)^k = frac {a^k}{b^k}$ to realize that that means $(frac ab)^{-k} = frac 1{(frac ab)^k}= frac 1{(frac {a^k}{b^k})} = frac {b^k}{a^k}$.

(Also $(frac ab)^{-k} = [(frac ab)^{-1}]^k = (frac ba)^k=frac {b^k}{a^k}$ or that $(frac ab)^{-k} = frac {a^{-k}}{b^{-k}} = frac {(frac 1{a^k})}{(frac 1{b^k})} = frac {b^k}{a^k}$.)

In any event

$(frac {2}{sqrt 6})^{-2} = (frac {sqrt 6}{ 2})^2 = frac {sqrt 6^2}{2^2} = frac 64 = frac 32$.

share|cite|improve this answer

answered 8 hours ago

fleabloodfleablood

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$endgroup$

share|cite|improve this answer

answered 8 hours ago

fleabloodfleablood

75.5k22 gold badges2828 silver badges9494 bronze badges

answered 8 hours ago

fleabloodfleablood

75.5k22 gold badges2828 silver badges9494 bronze badges

answered 8 hours ago

fleabloodfleablood

75.5k22 gold badges2828 silver badges9494 bronze badges

6

$begingroup$

$$begin{align}
-2ln left( frac{2}{sqrt 6} right) &= -2big( ln(2)-ln(sqrt{6}) big) \
&= -2ln(2)+2ln(6^{1/2}) \
&= -2ln(2)+ln(2cdot 3) \
&=-2ln(2)+ big( ln(2)+ln(3)big) \
&=ln(3)-ln(2)
end{align}$$
And for your specific question, remember that
$$left( frac{a}{b} right)^{-n}=left( frac{b}{a} right)^n$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

$endgroup$

add a comment | 

6

$begingroup$

$$begin{align}
-2ln left( frac{2}{sqrt 6} right) &= -2big( ln(2)-ln(sqrt{6}) big) \
&= -2ln(2)+2ln(6^{1/2}) \
&= -2ln(2)+ln(2cdot 3) \
&=-2ln(2)+ big( ln(2)+ln(3)big) \
&=ln(3)-ln(2)
end{align}$$
And for your specific question, remember that
$$left( frac{a}{b} right)^{-n}=left( frac{b}{a} right)^n$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

$endgroup$

add a comment | 

$begingroup$

$$begin{align}
-2ln left( frac{2}{sqrt 6} right) &= -2big( ln(2)-ln(sqrt{6}) big) \
&= -2ln(2)+2ln(6^{1/2}) \
&= -2ln(2)+ln(2cdot 3) \
&=-2ln(2)+ big( ln(2)+ln(3)big) \
&=ln(3)-ln(2)
end{align}$$
And for your specific question, remember that
$$left( frac{a}{b} right)^{-n}=left( frac{b}{a} right)^n$$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

$endgroup$

share|cite|improve this answer

edited 9 hours ago

answered 9 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

answered 9 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

answered 9 hours ago

Azif00Azif00

1,84911 silver badge1111 bronze badges

1

$begingroup$

Using your first step,

$-2 ln(frac{2}{sqrt{6}}) = ln(frac{2}{sqrt{6}})^{-2} = ln frac{1}{(frac{2}{sqrt{6}})^{2}} = ln frac{6}{4} = lnfrac{3}{2} = ln 3 - ln 2$

share|cite|improve this answer

answered 8 hours ago

Jaime Grimal AlvesJaime Grimal Alves

23877 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Using your first step,

$-2 ln(frac{2}{sqrt{6}}) = ln(frac{2}{sqrt{6}})^{-2} = ln frac{1}{(frac{2}{sqrt{6}})^{2}} = ln frac{6}{4} = lnfrac{3}{2} = ln 3 - ln 2$

share|cite|improve this answer

answered 8 hours ago

Jaime Grimal AlvesJaime Grimal Alves

23877 bronze badges

$endgroup$

add a comment | 

$begingroup$

Using your first step,

$-2 ln(frac{2}{sqrt{6}}) = ln(frac{2}{sqrt{6}})^{-2} = ln frac{1}{(frac{2}{sqrt{6}})^{2}} = ln frac{6}{4} = lnfrac{3}{2} = ln 3 - ln 2$

share|cite|improve this answer

answered 8 hours ago

Jaime Grimal AlvesJaime Grimal Alves

23877 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Jaime Grimal AlvesJaime Grimal Alves

23877 bronze badges

answered 8 hours ago

Jaime Grimal AlvesJaime Grimal Alves

23877 bronze badges

answered 8 hours ago

Jaime Grimal AlvesJaime Grimal Alves

23877 bronze badges

1

$begingroup$

Well you may start by distributing the index since $2$ and $sqrt 6$ are positive. Thus $$left(frac{2}{sqrt{6}}right)^{-2}=frac{2^{-2}}{{sqrt 6}^{-2}}.$$

Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^{-n}=frac {1}{a^n}.$$ Applying this to your expression, we have $$frac{2^{-2}}{{sqrt 6}^{-2}}=frac{frac {1}{2^2}}{frac{1}{{sqrt 6}^2}}=frac{frac {1}{4}}{frac{1}{6}}=frac{6}{4}=frac 32.$$

share|cite|improve this answer

answered 8 hours ago

AllawonderAllawonder

3,11788 silver badges1818 bronze badges

$endgroup$

add a comment | 

1

$begingroup$

Well you may start by distributing the index since $2$ and $sqrt 6$ are positive. Thus $$left(frac{2}{sqrt{6}}right)^{-2}=frac{2^{-2}}{{sqrt 6}^{-2}}.$$

Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^{-n}=frac {1}{a^n}.$$ Applying this to your expression, we have $$frac{2^{-2}}{{sqrt 6}^{-2}}=frac{frac {1}{2^2}}{frac{1}{{sqrt 6}^2}}=frac{frac {1}{4}}{frac{1}{6}}=frac{6}{4}=frac 32.$$

share|cite|improve this answer

answered 8 hours ago

AllawonderAllawonder

3,11788 silver badges1818 bronze badges

$endgroup$

add a comment | 

$begingroup$

Well you may start by distributing the index since $2$ and $sqrt 6$ are positive. Thus $$left(frac{2}{sqrt{6}}right)^{-2}=frac{2^{-2}}{{sqrt 6}^{-2}}.$$

Then recall that for any nonzero number $a$ and any negative integer $-n,$ we have $$a^{-n}=frac {1}{a^n}.$$ Applying this to your expression, we have $$frac{2^{-2}}{{sqrt 6}^{-2}}=frac{frac {1}{2^2}}{frac{1}{{sqrt 6}^2}}=frac{frac {1}{4}}{frac{1}{6}}=frac{6}{4}=frac 32.$$

share|cite|improve this answer

answered 8 hours ago

AllawonderAllawonder

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$endgroup$

share|cite|improve this answer

answered 8 hours ago

AllawonderAllawonder

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answered 8 hours ago

AllawonderAllawonder

3,11788 silver badges1818 bronze badges

answered 8 hours ago

AllawonderAllawonder

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0

$begingroup$

$$ -2 ln left( frac{2}{sqrt{6}} right) = ln 3 - ln 2$$

if and only if

$$ ln left( frac{2}{sqrt{6}} right)^{-2} = ln left( frac{3}{2}right) $$

if and only if

$$ ln left[ frac{1}{left(frac{2}{sqrt{6}}right)^{2}} right] = ln left( frac{3}{2}right)$$

And so, we have

$$ ln left( frac{6}{4} right) = ln left( frac{3}{2}right)$$

which is true.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

mlchristiansmlchristians

99422 silver badges1616 bronze badges

$endgroup$

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  you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse
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  – peek-a-boo
  9 hours ago
  

  
  
  
  


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  One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to include them initially.
  $endgroup$
  – mlchristians
  8 hours ago
  
  
  

  
  
  
  


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  $4 = 10 implies 7-3 =7+3 implies-3= 3 implies (-3)^2 = 3^2 implies 9=9$ which is true. Therefore $4 = 10$. Or if penguins were elephants then they would both eat food. Which is true. So penguins are elephants.
  $endgroup$
  – fleablood
  8 hours ago
  

  
  
  
  


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  "One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained." But if you do that you MUST specify the steps are actually if and only if steps. If you do not SPECIFICALLY state that it is reasonable to assume they are only forward implications and the proof is not valid.
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  – fleablood
  8 hours ago
  

  
  
  
  


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  Your first comment is not if and only if; and neither is you second. Forward implications is one-way; necessary and sufficient (i.e., if and only if are both ways). In the case of this problem, start from the given equation and work it to where I finished. Any problem with that. Now, if you want to do the work, start w/ the last statement and by a series of implications, show it implies the initial equation. I chose iff and only if instead.
  $endgroup$
  – mlchristians
  8 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

$$ -2 ln left( frac{2}{sqrt{6}} right) = ln 3 - ln 2$$

if and only if

$$ ln left( frac{2}{sqrt{6}} right)^{-2} = ln left( frac{3}{2}right) $$

if and only if

$$ ln left[ frac{1}{left(frac{2}{sqrt{6}}right)^{2}} right] = ln left( frac{3}{2}right)$$

And so, we have

$$ ln left( frac{6}{4} right) = ln left( frac{3}{2}right)$$

which is true.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

mlchristiansmlchristians

99422 silver badges1616 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you started by assuming the statement to be proven, so in reality you haven't shown anything. You need to run the argument in reverse
  $endgroup$
  – peek-a-boo
  9 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained. I neglected to include them initially.
  $endgroup$
  – mlchristians
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $4 = 10 implies 7-3 =7+3 implies-3= 3 implies (-3)^2 = 3^2 implies 9=9$ which is true. Therefore $4 = 10$. Or if penguins were elephants then they would both eat food. Which is true. So penguins are elephants.
  $endgroup$
  – fleablood
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  "One way to prove something is to take the LHS = RHS and proceed with a series of iff statements until something obvious is obtained." But if you do that you MUST specify the steps are actually if and only if steps. If you do not SPECIFICALLY state that it is reasonable to assume they are only forward implications and the proof is not valid.
  $endgroup$
  – fleablood
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your first comment is not if and only if; and neither is you second. Forward implications is one-way; necessary and sufficient (i.e., if and only if are both ways). In the case of this problem, start from the given equation and work it to where I finished. Any problem with that. Now, if you want to do the work, start w/ the last statement and by a series of implications, show it implies the initial equation. I chose iff and only if instead.
  $endgroup$
  – mlchristians
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

$$ -2 ln left( frac{2}{sqrt{6}} right) = ln 3 - ln 2$$

if and only if

$$ ln left( frac{2}{sqrt{6}} right)^{-2} = ln left( frac{3}{2}right) $$

if and only if

$$ ln left[ frac{1}{left(frac{2}{sqrt{6}}right)^{2}} right] = ln left( frac{3}{2}right)$$

And so, we have

$$ ln left( frac{6}{4} right) = ln left( frac{3}{2}right)$$

which is true.

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

mlchristiansmlchristians

99422 silver badges1616 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 9 hours ago

mlchristiansmlchristians

99422 silver badges1616 bronze badges

answered 9 hours ago

mlchristiansmlchristians

99422 silver badges1616 bronze badges

answered 9 hours ago

mlchristiansmlchristians

99422 silver badges1616 bronze badges