Ggthjy

What do *foreign films* mean for an American?

Should I take out a loan for a friend to invest on my behalf?

Why aren't there more Gauls like Obelix?

PTIJ: Why does only a Shor Tam ask at the Seder, and not a Shor Mu'ad?

Are there historical instances of the capital of a colonising country being temporarily or permanently shifted to one of its colonies?

Finitely many repeated replacements

I can't die. Who am I?

What materials can be used to make a humanoid skin warm?

Windows Server Datacenter Edition - Unlimited Virtual Machines

Can't make sense of a paragraph from Lovecraft

How many characters using PHB rules does it take to be able to have access to any PHB spell at the start of an adventuring day?

Why do phishing e-mails use faked e-mail addresses instead of the real one?

Why couldn't the separatists legally leave the Republic?

Can the alpha, lambda values of a glmnet object output determine whether ridge or Lasso?

Would an aboleth's Phantasmal Force lair action be affected by Counterspell, Dispel Magic, and/or Slow?

Are all players supposed to be able to see each others' character sheets?

Why restrict private health insurance?

Source permutation

Which situations would cause a company to ground or recall a aircraft series?

What is better: yes / no radio, or simple checkbox?

How exactly does an Ethernet collision happen in the cable, since nodes use different circuits for Tx and Rx?

Called into a meeting and told we are being made redundant (laid off) and "not to share outside". Can I tell my partner?

Are small insurances worth it?

Did Amazon pay $0 in taxes last year?

Help find my computational error for logarithms

What are logarithms, and what do they do?Difference of Logarithms to form a quotient?simple logarithms with exponentsSimple question on logarithmsSolving exponential equations using logarithmsI need help identifying the slope for an equation.Solutions of $2^x 7^{1/x}le 14$Adding logarithms with different basesSolving Logarithms HelpHow do you multiply these logarithms and find the domain?

1

1

$begingroup$

I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?

algebra-precalculus logarithms

share|cite|improve this question

edited 24 mins ago

Eevee Trainer

7,70521338

asked 30 mins ago

KevinKevin

396

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac 1 {log x}$ is not generally the same as $log frac 1 x $
  $endgroup$
  – J. W. Tanner
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  11 mins ago
  
  
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?

algebra-precalculus logarithms

share|cite|improve this question

edited 24 mins ago

Eevee Trainer

7,70521338

asked 30 mins ago

KevinKevin

396

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac 1 {log x}$ is not generally the same as $log frac 1 x $
  $endgroup$
  – J. W. Tanner
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  11 mins ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I am supposed to get $x = 8$ and $x = x^{-2/3}$. What did I do wrong?

algebra-precalculus logarithms

share|cite|improve this question

edited 24 mins ago

Eevee Trainer

7,70521338

asked 30 mins ago

KevinKevin

396

$endgroup$

share|cite|improve this question

edited 24 mins ago

Eevee Trainer

7,70521338

asked 30 mins ago

KevinKevin

396

share|cite|improve this question

edited 24 mins ago

Eevee Trainer

7,70521338

asked 30 mins ago

KevinKevin

396

edited 24 mins ago

Eevee Trainer

7,70521338

edited 24 mins ago

Eevee Trainer

7,70521338

asked 30 mins ago

KevinKevin

396

asked 30 mins ago

KevinKevin

396

2 Answers
2

active

oldest

votes

2

$begingroup$

Note that

$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$

For example, take $x = 4$. Then

$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$

but

$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$

This is where your error lies.

share|cite|improve this answer

answered 27 mins ago

Eevee TrainerEevee Trainer

7,70521338

$endgroup$

add a comment | 

2

$begingroup$

To get the correct answer, let $L=log_2(x).$

Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$

Multiply by $6L$ to get $$3L^2-6=7L.$$

Thus $$3L^2-7L-6=0$$

or $$(3L+2)(L-3)=0.$$

Can you take it from here?

share|cite|improve this answer

answered 21 mins ago

J. W. TannerJ. W. Tanner

3,0501320

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
  $endgroup$
  – Eevee Trainer
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
  $endgroup$
  – Kevin
  1 min ago
  

  
  
  
  

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%2f3143253%2fhelp-find-my-computational-error-for-logarithms%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Note that

$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$

For example, take $x = 4$. Then

$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$

but

$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$

This is where your error lies.

share|cite|improve this answer

answered 27 mins ago

Eevee TrainerEevee Trainer

7,70521338

$endgroup$

add a comment | 

2

$begingroup$

Note that

$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$

For example, take $x = 4$. Then

$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$

but

$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$

This is where your error lies.

share|cite|improve this answer

answered 27 mins ago

Eevee TrainerEevee Trainer

7,70521338

$endgroup$

add a comment | 

$begingroup$

Note that

$$left( log_2(x) right)^{-1} neq log_2 left(x^{-1} right)$$

For example, take $x = 4$. Then

$$left( log_2(x) right)^{-1} = left( log_2(4) right)^{-1} = 2^{-1} = frac 1 2$$

but

$$log_2 left(x^{-1} right) = log_2 left( frac 1 4 right) = -2$$

This is where your error lies.

share|cite|improve this answer

answered 27 mins ago

Eevee TrainerEevee Trainer

7,70521338

$endgroup$

share|cite|improve this answer

answered 27 mins ago

Eevee TrainerEevee Trainer

7,70521338

answered 27 mins ago

Eevee TrainerEevee Trainer

7,70521338

answered 27 mins ago

Eevee TrainerEevee Trainer

7,70521338

2

$begingroup$

To get the correct answer, let $L=log_2(x).$

Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$

Multiply by $6L$ to get $$3L^2-6=7L.$$

Thus $$3L^2-7L-6=0$$

or $$(3L+2)(L-3)=0.$$

Can you take it from here?

share|cite|improve this answer

answered 21 mins ago

J. W. TannerJ. W. Tanner

3,0501320

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
  $endgroup$
  – Eevee Trainer
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
  $endgroup$
  – Kevin
  1 min ago
  

  
  
  
  

add a comment | 

2

$begingroup$

To get the correct answer, let $L=log_2(x).$

Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$

Multiply by $6L$ to get $$3L^2-6=7L.$$

Thus $$3L^2-7L-6=0$$

or $$(3L+2)(L-3)=0.$$

Can you take it from here?

share|cite|improve this answer

answered 21 mins ago

J. W. TannerJ. W. Tanner

3,0501320

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Not gonna lie, that's a pretty slick solution (even if it's just a minor trick involving a substitution - I always love it when something like that makes the work look 10 times easier :p).
  $endgroup$
  – Eevee Trainer
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know how to solve in this way, I was just confused about why I got a different answer when using another approach. But still thank you for your help ;)
  $endgroup$
  – Kevin
  1 min ago
  

  
  
  
  

add a comment | 

$begingroup$

To get the correct answer, let $L=log_2(x).$

Then we have $$frac 1 2 L - frac 1 L = frac 7 6.$$

Multiply by $6L$ to get $$3L^2-6=7L.$$

Thus $$3L^2-7L-6=0$$

or $$(3L+2)(L-3)=0.$$

Can you take it from here?

share|cite|improve this answer

answered 21 mins ago

J. W. TannerJ. W. Tanner

3,0501320

$endgroup$

share|cite|improve this answer

answered 21 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 21 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 21 mins ago

J. W. TannerJ. W. Tanner

3,0501320