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Help find my computational error for logarithms

Solving for Exponents and LogarithmsAlgebra help, logarithms, calculusReal life applications for logarithmsLaws of Logarithms IssueGraphing natural logarithmsusing logarithms to solve the following equation to find xSolutions of $2^x 7^{1/x}le 14$Natural Logarithms (Help)Adding logarithms with different basesFind a simplified expression without logarithms

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 28 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  16 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 28 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  16 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 28 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

$endgroup$

share|cite|improve this question

edited 28 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

share|cite|improve this question

edited 28 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

edited 28 mins ago

Eevee Trainer

7,71521338

edited 28 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

asked 34 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 31 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 25 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
  19 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
  5 mins ago
  

  
  
  
  

add a comment | 

Your Answer

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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 31 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 31 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 31 mins ago

Eevee TrainerEevee Trainer

7,71521338

$endgroup$

share|cite|improve this answer

answered 31 mins ago

Eevee TrainerEevee Trainer

7,71521338

answered 31 mins ago

Eevee TrainerEevee Trainer

7,71521338

answered 31 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 25 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
  19 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
  5 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 25 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
  19 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
  5 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 25 mins ago

J. W. TannerJ. W. Tanner

3,0501320

$endgroup$

share|cite|improve this answer

answered 25 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 25 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 25 mins ago

J. W. TannerJ. W. Tanner

3,0501320