Gsyfuy

What do you call someone who likes to pick fights?

Haman going to the second feast dirty

Why is a very small peak with larger m/z not considered to be the molecular ion?

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

What problems would a superhuman have who's skin is constantly hot?

How can I get players to focus on the story aspect of D&D?

Was it really inappropriate to write a pull request for the company I interviewed with?

Do I really need to have a scientific explanation for my premise?

Is a piano played in the same way as a harmonium?

Is it possible to find 2014 distinct positive integers whose sum is divisible by each of them?

In the late 1940’s to early 1950’s what technology was available that could melt a LOT of ice?

Can I negotiate a patent idea for a raise, under French law?

Street obstacles in New Zealand

Finitely many repeated replacements

Outlet with 3 sets of wires

Why aren't there more Gauls like Obelix?

Which classes are needed to have access to every spell in the PHB?

Why couldn't the separatists legally leave the Republic?

When Schnorr signatures are part of Bitcoin will it be possible validate each block with only one signature validation?

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

What is this diamond of every day?

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

Confusion about Complex Continued Fraction

Signed and unsigned numbers

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 33 mins ago

Eevee Trainer

7,71521338

asked 39 mins ago

KevinKevin

396

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  21 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 33 mins ago

Eevee Trainer

7,71521338

asked 39 mins ago

KevinKevin

396

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  21 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 33 mins ago

Eevee Trainer

7,71521338

asked 39 mins ago

KevinKevin

396

$endgroup$

share|cite|improve this question

edited 33 mins ago

Eevee Trainer

7,71521338

asked 39 mins ago

KevinKevin

396

share|cite|improve this question

edited 33 mins ago

Eevee Trainer

7,71521338

asked 39 mins ago

KevinKevin

396

edited 33 mins ago

Eevee Trainer

7,71521338

edited 33 mins ago

Eevee Trainer

7,71521338

asked 39 mins ago

KevinKevin

396

asked 39 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 36 mins ago

Eevee TrainerEevee Trainer

7,71521338

$endgroup$

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 30 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
  24 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
  10 mins 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 36 mins ago

Eevee TrainerEevee Trainer

7,71521338

$endgroup$

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 36 mins ago

Eevee TrainerEevee Trainer

7,71521338

$endgroup$

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 36 mins ago

Eevee TrainerEevee Trainer

7,71521338

$endgroup$

share|cite|improve this answer

answered 36 mins ago

Eevee TrainerEevee Trainer

7,71521338

answered 36 mins ago

Eevee TrainerEevee Trainer

7,71521338

answered 36 mins ago

Eevee TrainerEevee Trainer

7,71521338

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 30 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
  24 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
  10 mins 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 30 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
  24 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
  10 mins 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 30 mins ago

J. W. TannerJ. W. Tanner

3,0501320

$endgroup$

share|cite|improve this answer

answered 30 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 30 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 30 mins ago

J. W. TannerJ. W. Tanner

3,0501320