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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 27 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
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  15 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 27 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
  30 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^{-2/3}$ ?
  $endgroup$
  – J. W. Tanner
  15 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 27 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

$endgroup$

share|cite|improve this question

edited 27 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

share|cite|improve this question

edited 27 mins ago

Eevee Trainer

7,71521338

asked 34 mins ago

KevinKevin

396

edited 27 mins ago

Eevee Trainer

7,71521338

edited 27 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 30 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 24 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
  18 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
  4 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 30 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 30 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 30 mins ago

Eevee TrainerEevee Trainer

7,71521338

$endgroup$

share|cite|improve this answer

answered 30 mins ago

Eevee TrainerEevee Trainer

7,71521338

answered 30 mins ago

Eevee TrainerEevee Trainer

7,71521338

answered 30 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 24 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
  18 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
  4 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 24 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
  18 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
  4 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 24 mins ago

J. W. TannerJ. W. Tanner

3,0501320

$endgroup$

share|cite|improve this answer

answered 24 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 24 mins ago

J. W. TannerJ. W. Tanner

3,0501320

answered 24 mins ago

J. W. TannerJ. W. Tanner

3,0501320