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Why is $frac{13}{10} = e^{lnfrac{10}{13}*x}$

Canceling in fractions sometimes gives a wrong resultHow to solve this linear equation? which has an x on each sideWhy isn't $frac{0^{0!}}{0!^0}$ not undefined?Why is $lim_{x to 0} x = (1+tan x)^frac{1}{x}$ not 1?Why is $sin^{-1}(sin(frac{5pi}{8}))ne frac{5pi}{8}$?Calculating $x$ from two equations involving $e^x$ and why is $ln{(frac{0.5}{0.1})}$ not the same as $frac{ln{0.5}}{ln{0.1}}$Why is $frac{log 8}{log 3} = frac{ln8}{ln3}$$frac{4}{5^{x+1}}-frac{1}{5^x} = -0.04$Why is $frac{10xDelta x +5( Delta x)^2}{Delta x}=10x+ 5Delta x$? (Algebra error)Domain of exponential function

2

$begingroup$

Why is:
$$frac{13}{10}^{-x} = e^{lnfrac{10}{13}*x}$ $

I thought that the $ln$ and the $e$ canceled out making it equivalent to saying:
$$e^{lnfrac{10}{13}*x} = frac{10x}{13}$$

Is this not correct?

algebra-precalculus exponential-function

share|cite|improve this question

edited Mar 15 at 1:45

DavidG

1

asked Mar 12 at 4:57

John RawlsJohn Rawls

1,281619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Eevee Trainer: You have assumed where the parentheses should go, which OP did not specify. The last line is not correct and I can't tell if that is the intent. I would roll back and let OP clarify.
  $endgroup$
  – Ross Millikan
  Mar 12 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The positioning of the parentheses is both immediately clear from the nature of the problem and the formatting in the original code, frac{13}{10}^{-x}, at least in my opinion. The last line I would argue is a simple mistake on the OP's behalf owing to some misconceptions about how $e$ and the natural logarithm interact. But as you wish.
  $endgroup$
  – Eevee Trainer
  Mar 12 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer: I am more persnickety than many about parentheses. We see lots of posts where they are omitted, sometimes it is clear what is meant and sometimes not. If I see $1/2x$ is that $frac 1{2x}$ or $frac 12x$? I then err on the side of forcing OP to answer.
  $endgroup$
  – Ross Millikan
  Mar 12 at 5:18
  

  
  
  
  

add a comment | 

2

$begingroup$

Why is:
$$frac{13}{10}^{-x} = e^{lnfrac{10}{13}*x}$ $

I thought that the $ln$ and the $e$ canceled out making it equivalent to saying:
$$e^{lnfrac{10}{13}*x} = frac{10x}{13}$$

Is this not correct?

algebra-precalculus exponential-function

share|cite|improve this question

edited Mar 15 at 1:45

DavidG

1

asked Mar 12 at 4:57

John RawlsJohn Rawls

1,281619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Eevee Trainer: You have assumed where the parentheses should go, which OP did not specify. The last line is not correct and I can't tell if that is the intent. I would roll back and let OP clarify.
  $endgroup$
  – Ross Millikan
  Mar 12 at 5:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The positioning of the parentheses is both immediately clear from the nature of the problem and the formatting in the original code, frac{13}{10}^{-x}, at least in my opinion. The last line I would argue is a simple mistake on the OP's behalf owing to some misconceptions about how $e$ and the natural logarithm interact. But as you wish.
  $endgroup$
  – Eevee Trainer
  Mar 12 at 5:12
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer: I am more persnickety than many about parentheses. We see lots of posts where they are omitted, sometimes it is clear what is meant and sometimes not. If I see $1/2x$ is that $frac 1{2x}$ or $frac 12x$? I then err on the side of forcing OP to answer.
  $endgroup$
  – Ross Millikan
  Mar 12 at 5:18
  

  
  
  
  

add a comment | 

$begingroup$

Why is:
$$frac{13}{10}^{-x} = e^{lnfrac{10}{13}*x}$ $

I thought that the $ln$ and the $e$ canceled out making it equivalent to saying:
$$e^{lnfrac{10}{13}*x} = frac{10x}{13}$$

Is this not correct?

algebra-precalculus exponential-function

share|cite|improve this question

edited Mar 15 at 1:45

DavidG

1

asked Mar 12 at 4:57

John RawlsJohn Rawls

1,281619

$endgroup$

share|cite|improve this question

edited Mar 15 at 1:45

DavidG

1

asked Mar 12 at 4:57

John RawlsJohn Rawls

1,281619

share|cite|improve this question

edited Mar 15 at 1:45

DavidG

1

asked Mar 12 at 4:57

John RawlsJohn Rawls

1,281619

edited Mar 15 at 1:45

DavidG

1

edited Mar 15 at 1:45

DavidG

1

asked Mar 12 at 4:57

John RawlsJohn Rawls

1,281619

asked Mar 12 at 4:57

John RawlsJohn Rawls

1,281619

5 Answers
5

active

oldest

votes

2

$begingroup$

$$
e^{xln(10/13)}
=e^{ln((10/13)^{x})}
=left(frac{10}{13}right)^{x}
=left(frac{13}{10}right)^{-x}
$$

share|cite|improve this answer

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

$endgroup$

add a comment | 

2

$begingroup$

Your question is missing the $-x$ term in the power.

Further, $ln$ never canceled out with $e$, it is by the property: $e^{ln x}=x$

Now in your case, $frac{13}{10}^{-x}=left(left(frac{13}{10}right)^{-1}right)^x=left(frac{10}{13}right)^x=e^{xln (frac{10}{13})}$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,528519

$endgroup$

add a comment | 

1

$begingroup$

Notice that the fractions in each case are flipped, and that taking the reciprocal of a fraction is the same as raising it to the power of $-1$. Then we have

$$e^{ln(10/13) cdot x} = left(e^{ln(10/13)} right)^x= left( frac{10}{13} right)^x = left( left( frac{13}{10} right)^{-1} right)^x = left( frac{13}{10} right)^{-x}$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,06921439

$endgroup$

add a comment | 

0

$begingroup$

You need some parentheses to show the order of operations. You title does not match the first line of the question, but the question looks to match what is going on here. When you write $frac {13}{10}^{-x}$ do you mean $left(frac {13}{10}right)^{-x}$ or $frac {(13^{-x})}{10}$? Similarly, when you write $ e^{lnfrac{10}{13}*x}$ do you mean $ e^{(lnfrac{10}{13})*x}$ or $ e^{ln(frac{10}{13}*x)}$?

For the question as edited (is that what you meant?) you can do
$$left( frac{13}{10} right)^{-x} =left(e^{ln frac {13}{10}}right)^{-x} \
=left(e^{-xln frac {13}{10}}right)\
=e^{xln(10/13)}$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

299k24200374

$endgroup$

add a comment | 

0

$begingroup$

You are correct that:

$e^{ln (frac{10}{13}x)}=frac {10}{13}x $.

But

$e^{(ln frac {10}{13})x}=(e^{ln frac {10}{13}})^x=(frac {10}{13})^x=(frac {13}{10})^{-x} $.

You aren't missing any concept. You are just not reading the question as the person who wrote it intended. Which, arguably, could be the authors fault.

share|cite|improve this answer

answered Mar 12 at 6:20

fleabloodfleablood

72.8k22788

$endgroup$

add a comment | 

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2

$begingroup$

$$
e^{xln(10/13)}
=e^{ln((10/13)^{x})}
=left(frac{10}{13}right)^{x}
=left(frac{13}{10}right)^{-x}
$$

share|cite|improve this answer

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

$endgroup$

add a comment | 

2

$begingroup$

$$
e^{xln(10/13)}
=e^{ln((10/13)^{x})}
=left(frac{10}{13}right)^{x}
=left(frac{13}{10}right)^{-x}
$$

share|cite|improve this answer

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

$endgroup$

add a comment | 

$begingroup$

$$
e^{xln(10/13)}
=e^{ln((10/13)^{x})}
=left(frac{10}{13}right)^{x}
=left(frac{13}{10}right)^{-x}
$$

share|cite|improve this answer

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

$endgroup$

share|cite|improve this answer

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

2

$begingroup$

Your question is missing the $-x$ term in the power.

Further, $ln$ never canceled out with $e$, it is by the property: $e^{ln x}=x$

Now in your case, $frac{13}{10}^{-x}=left(left(frac{13}{10}right)^{-1}right)^x=left(frac{10}{13}right)^x=e^{xln (frac{10}{13})}$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,528519

$endgroup$

add a comment | 

2

$begingroup$

Your question is missing the $-x$ term in the power.

Further, $ln$ never canceled out with $e$, it is by the property: $e^{ln x}=x$

Now in your case, $frac{13}{10}^{-x}=left(left(frac{13}{10}right)^{-1}right)^x=left(frac{10}{13}right)^x=e^{xln (frac{10}{13})}$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,528519

$endgroup$

add a comment | 

$begingroup$

Your question is missing the $-x$ term in the power.

Further, $ln$ never canceled out with $e$, it is by the property: $e^{ln x}=x$

Now in your case, $frac{13}{10}^{-x}=left(left(frac{13}{10}right)^{-1}right)^x=left(frac{10}{13}right)^x=e^{xln (frac{10}{13})}$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,528519

$endgroup$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,528519

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,528519

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,528519

1

$begingroup$

Notice that the fractions in each case are flipped, and that taking the reciprocal of a fraction is the same as raising it to the power of $-1$. Then we have

$$e^{ln(10/13) cdot x} = left(e^{ln(10/13)} right)^x= left( frac{10}{13} right)^x = left( left( frac{13}{10} right)^{-1} right)^x = left( frac{13}{10} right)^{-x}$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,06921439

$endgroup$

add a comment | 

1

$begingroup$

Notice that the fractions in each case are flipped, and that taking the reciprocal of a fraction is the same as raising it to the power of $-1$. Then we have

$$e^{ln(10/13) cdot x} = left(e^{ln(10/13)} right)^x= left( frac{10}{13} right)^x = left( left( frac{13}{10} right)^{-1} right)^x = left( frac{13}{10} right)^{-x}$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,06921439

$endgroup$

add a comment | 

$begingroup$

Notice that the fractions in each case are flipped, and that taking the reciprocal of a fraction is the same as raising it to the power of $-1$. Then we have

$$e^{ln(10/13) cdot x} = left(e^{ln(10/13)} right)^x= left( frac{10}{13} right)^x = left( left( frac{13}{10} right)^{-1} right)^x = left( frac{13}{10} right)^{-x}$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,06921439

$endgroup$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,06921439

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,06921439

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,06921439

0

$begingroup$

You need some parentheses to show the order of operations. You title does not match the first line of the question, but the question looks to match what is going on here. When you write $frac {13}{10}^{-x}$ do you mean $left(frac {13}{10}right)^{-x}$ or $frac {(13^{-x})}{10}$? Similarly, when you write $ e^{lnfrac{10}{13}*x}$ do you mean $ e^{(lnfrac{10}{13})*x}$ or $ e^{ln(frac{10}{13}*x)}$?

For the question as edited (is that what you meant?) you can do
$$left( frac{13}{10} right)^{-x} =left(e^{ln frac {13}{10}}right)^{-x} \
=left(e^{-xln frac {13}{10}}right)\
=e^{xln(10/13)}$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

299k24200374

$endgroup$

add a comment | 

0

$begingroup$

You need some parentheses to show the order of operations. You title does not match the first line of the question, but the question looks to match what is going on here. When you write $frac {13}{10}^{-x}$ do you mean $left(frac {13}{10}right)^{-x}$ or $frac {(13^{-x})}{10}$? Similarly, when you write $ e^{lnfrac{10}{13}*x}$ do you mean $ e^{(lnfrac{10}{13})*x}$ or $ e^{ln(frac{10}{13}*x)}$?

For the question as edited (is that what you meant?) you can do
$$left( frac{13}{10} right)^{-x} =left(e^{ln frac {13}{10}}right)^{-x} \
=left(e^{-xln frac {13}{10}}right)\
=e^{xln(10/13)}$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

299k24200374

$endgroup$

add a comment | 

$begingroup$

You need some parentheses to show the order of operations. You title does not match the first line of the question, but the question looks to match what is going on here. When you write $frac {13}{10}^{-x}$ do you mean $left(frac {13}{10}right)^{-x}$ or $frac {(13^{-x})}{10}$? Similarly, when you write $ e^{lnfrac{10}{13}*x}$ do you mean $ e^{(lnfrac{10}{13})*x}$ or $ e^{ln(frac{10}{13}*x)}$?

For the question as edited (is that what you meant?) you can do
$$left( frac{13}{10} right)^{-x} =left(e^{ln frac {13}{10}}right)^{-x} \
=left(e^{-xln frac {13}{10}}right)\
=e^{xln(10/13)}$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

299k24200374

$endgroup$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

299k24200374

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

299k24200374

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

299k24200374

0

$begingroup$

You are correct that:

$e^{ln (frac{10}{13}x)}=frac {10}{13}x $.

But

$e^{(ln frac {10}{13})x}=(e^{ln frac {10}{13}})^x=(frac {10}{13})^x=(frac {13}{10})^{-x} $.

You aren't missing any concept. You are just not reading the question as the person who wrote it intended. Which, arguably, could be the authors fault.

share|cite|improve this answer

answered Mar 12 at 6:20

fleabloodfleablood

72.8k22788

$endgroup$

add a comment | 

0

$begingroup$

You are correct that:

$e^{ln (frac{10}{13}x)}=frac {10}{13}x $.

But

$e^{(ln frac {10}{13})x}=(e^{ln frac {10}{13}})^x=(frac {10}{13})^x=(frac {13}{10})^{-x} $.

You aren't missing any concept. You are just not reading the question as the person who wrote it intended. Which, arguably, could be the authors fault.

share|cite|improve this answer

answered Mar 12 at 6:20

fleabloodfleablood

72.8k22788

$endgroup$

add a comment | 

$begingroup$

You are correct that:

$e^{ln (frac{10}{13}x)}=frac {10}{13}x $.

But

$e^{(ln frac {10}{13})x}=(e^{ln frac {10}{13}})^x=(frac {10}{13})^x=(frac {13}{10})^{-x} $.

You aren't missing any concept. You are just not reading the question as the person who wrote it intended. Which, arguably, could be the authors fault.

share|cite|improve this answer

answered Mar 12 at 6:20

fleabloodfleablood

72.8k22788

$endgroup$

share|cite|improve this answer

answered Mar 12 at 6:20

fleabloodfleablood

72.8k22788

answered Mar 12 at 6:20

fleabloodfleablood

72.8k22788

answered Mar 12 at 6:20

fleabloodfleablood

72.8k22788