Fdtxjr

PTIJ: Haman's bad computer

What does chmod -u do?

Why is this estimator biased?

Open a doc from terminal, but not by its name

Are Captain Marvel's powers affected by Thanos' actions in Infinity War

Creepy dinosaur pc game identification

What exact color does ozone gas have?

How to hide some fields of struct in C?

Does Doodling or Improvising on the Piano Have Any Benefits?

Pre-mixing cryogenic fuels and using only one fuel tank

How to fade a semiplane defined by line?

Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?

Why does the Sun have different day lengths, but not the gas giants?

Mimic lecturing on blackboard, facing audience

Store Credit Card Information in Password Manager?

Plot of a tornado-shaped surface

Hero deduces identity of a killer

Can I still be respawned if I die by falling off the map?

Why would a new[] expression ever invoke a destructor?

What is the highest possible scrabble score for placing a single tile

Calculating total slots

Quoting Keynes in a lecture

What if a revenant (monster) gains fire resistance?

Why does AES have exactly 10 rounds for a 128-bit key, 12 for 192 bits and 14 for a 256-bit key size?

Why is $frac1310 = e^lnfrac1013*x$

Canceling in fractions sometimes gives a wrong resultHow to solve this linear equation? which has an x on each sideWhy isn't $frac0^0!0!^0$ not undefined?Why is $lim_x to 0 x = (1+tan x)^frac1x$ not 1?Why is $sin^-1(sin(frac5pi8))ne frac5pi8$?Calculating $x$ from two equations involving $e^x$ and why is $ln(frac0.50.1)$ not the same as $fracln0.5ln0.1$Why is $fraclog 8log 3 = fracln8ln3$$frac45^x+1-frac15^x = -0.04$Why is $frac10xDelta x +5( Delta x)^2Delta x=10x+ 5Delta x$? (Algebra error)Domain of exponential function

2

$begingroup$

Why is:
$$frac1310^-x = e^lnfrac1013*x$ $

I thought that the $ln$ and the $e$ canceled out making it equivalent to saying:
$$e^lnfrac1013*x = frac10x13$$

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, frac1310^-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 $frac 12x$ 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:
$$frac1310^-x = e^lnfrac1013*x$ $

I thought that the $ln$ and the $e$ canceled out making it equivalent to saying:
$$e^lnfrac1013*x = frac10x13$$

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, frac1310^-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 $frac 12x$ 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:
$$frac1310^-x = e^lnfrac1013*x$ $

I thought that the $ln$ and the $e$ canceled out making it equivalent to saying:
$$e^lnfrac1013*x = frac10x13$$

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

$begingroup$

$$
e^xln(10/13)
=e^ln((10/13)^x)
=left(frac1013right)^x
=left(frac1310right)^-x
$$

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, $frac1310^-x=left(left(frac1310right)^-1right)^x=left(frac1013right)^x=e^xln (frac1013)$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,553519

$endgroup$

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

$$e^ln(10/13) cdot x = left(e^ln(10/13) right)^x= left( frac1013 right)^x = left( left( frac1310 right)^-1 right)^x = left( frac1310 right)^-x$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,38921439

$endgroup$

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 $frac 1310^-x$ do you mean $left(frac 1310right)^-x$ or $frac (13^-x)10$? Similarly, when you write $ e^lnfrac1013*x$ do you mean $ e^(lnfrac1013)*x$ or $ e^ln(frac1013*x)$?

For the question as edited (is that what you meant?) you can do
$$left( frac1310 right)^-x =left(e^ln frac 1310right)^-x \
=left(e^-xln frac 1310right)\
=e^xln(10/13)$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

300k24200374

$endgroup$

add a comment | 

0

$begingroup$

You are correct that:

$e^ln (frac1013x)=frac 1013x $.

But

$e^(ln frac 1013)x=(e^ln frac 1013)^x=(frac 1013)^x=(frac 1310)^-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

72.9k22789

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3144677%2fwhy-is-frac1310-e-ln-frac1013x%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

$$
e^xln(10/13)
=e^ln((10/13)^x)
=left(frac1013right)^x
=left(frac1310right)^-x
$$

share|cite|improve this answer

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

$endgroup$

add a comment | 

2

$begingroup$

$$
e^xln(10/13)
=e^ln((10/13)^x)
=left(frac1013right)^x
=left(frac1310right)^-x
$$

share|cite|improve this answer

answered Mar 12 at 5:01

parsiadparsiad

18.5k32453

$endgroup$

add a comment | 

$begingroup$

$$
e^xln(10/13)
=e^ln((10/13)^x)
=left(frac1013right)^x
=left(frac1310right)^-x
$$

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, $frac1310^-x=left(left(frac1310right)^-1right)^x=left(frac1013right)^x=e^xln (frac1013)$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,553519

$endgroup$

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, $frac1310^-x=left(left(frac1310right)^-1right)^x=left(frac1013right)^x=e^xln (frac1013)$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,553519

$endgroup$

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, $frac1310^-x=left(left(frac1310right)^-1right)^x=left(frac1013right)^x=e^xln (frac1013)$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,553519

$endgroup$

share|cite|improve this answer

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,553519

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,553519

answered Mar 12 at 5:05

Sujit BhattacharyyaSujit Bhattacharyya

1,553519

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

$$e^ln(10/13) cdot x = left(e^ln(10/13) right)^x= left( frac1013 right)^x = left( left( frac1310 right)^-1 right)^x = left( frac1310 right)^-x$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,38921439

$endgroup$

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

$$e^ln(10/13) cdot x = left(e^ln(10/13) right)^x= left( frac1013 right)^x = left( left( frac1310 right)^-1 right)^x = left( frac1310 right)^-x$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,38921439

$endgroup$

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

$$e^ln(10/13) cdot x = left(e^ln(10/13) right)^x= left( frac1013 right)^x = left( left( frac1310 right)^-1 right)^x = left( frac1310 right)^-x$$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,38921439

$endgroup$

share|cite|improve this answer

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,38921439

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,38921439

answered Mar 12 at 5:02

Eevee TrainerEevee Trainer

8,38921439

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 $frac 1310^-x$ do you mean $left(frac 1310right)^-x$ or $frac (13^-x)10$? Similarly, when you write $ e^lnfrac1013*x$ do you mean $ e^(lnfrac1013)*x$ or $ e^ln(frac1013*x)$?

For the question as edited (is that what you meant?) you can do
$$left( frac1310 right)^-x =left(e^ln frac 1310right)^-x \
=left(e^-xln frac 1310right)\
=e^xln(10/13)$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

300k24200374

$endgroup$

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 $frac 1310^-x$ do you mean $left(frac 1310right)^-x$ or $frac (13^-x)10$? Similarly, when you write $ e^lnfrac1013*x$ do you mean $ e^(lnfrac1013)*x$ or $ e^ln(frac1013*x)$?

For the question as edited (is that what you meant?) you can do
$$left( frac1310 right)^-x =left(e^ln frac 1310right)^-x \
=left(e^-xln frac 1310right)\
=e^xln(10/13)$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

300k24200374

$endgroup$

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 $frac 1310^-x$ do you mean $left(frac 1310right)^-x$ or $frac (13^-x)10$? Similarly, when you write $ e^lnfrac1013*x$ do you mean $ e^(lnfrac1013)*x$ or $ e^ln(frac1013*x)$?

For the question as edited (is that what you meant?) you can do
$$left( frac1310 right)^-x =left(e^ln frac 1310right)^-x \
=left(e^-xln frac 1310right)\
=e^xln(10/13)$$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

300k24200374

$endgroup$

share|cite|improve this answer

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

300k24200374

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

300k24200374

answered Mar 12 at 5:16

Ross MillikanRoss Millikan

300k24200374

0

$begingroup$

You are correct that:

$e^ln (frac1013x)=frac 1013x $.

But

$e^(ln frac 1013)x=(e^ln frac 1013)^x=(frac 1013)^x=(frac 1310)^-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

72.9k22789

$endgroup$

add a comment | 

0

$begingroup$

You are correct that:

$e^ln (frac1013x)=frac 1013x $.

But

$e^(ln frac 1013)x=(e^ln frac 1013)^x=(frac 1013)^x=(frac 1310)^-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

72.9k22789

$endgroup$

add a comment | 

$begingroup$

You are correct that:

$e^ln (frac1013x)=frac 1013x $.

But

$e^(ln frac 1013)x=(e^ln frac 1013)^x=(frac 1013)^x=(frac 1310)^-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

72.9k22789

$endgroup$

share|cite|improve this answer

answered Mar 12 at 6:20

fleabloodfleablood

72.9k22789

answered Mar 12 at 6:20

fleabloodfleablood

72.9k22789

answered Mar 12 at 6:20

fleabloodfleablood

72.9k22789