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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 $2^x 7^1/xle 14$Adding logarithms with different basesSolving Logarithms HelpHow do you multiply these logarithms and find the domain?

2

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 53 mins ago

Eevee Trainer

7,72621338

asked 59 mins ago

KevinKevin

446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

2

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 53 mins ago

Eevee Trainer

7,72621338

asked 59 mins ago

KevinKevin

446

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^-2/3$ ?
  $endgroup$
  – J. W. Tanner
  40 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 53 mins ago

Eevee Trainer

7,72621338

asked 59 mins ago

KevinKevin

446

$endgroup$

share|cite|improve this question

edited 53 mins ago

Eevee Trainer

7,72621338

asked 59 mins ago

KevinKevin

446

share|cite|improve this question

edited 53 mins ago

Eevee Trainer

7,72621338

asked 59 mins ago

KevinKevin

446

edited 53 mins ago

Eevee Trainer

7,72621338

edited 53 mins ago

Eevee Trainer

7,72621338

asked 59 mins ago

KevinKevin

446

asked 59 mins ago

KevinKevin

446

2 Answers
2

active

oldest

votes

3

$begingroup$

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 56 mins ago

Eevee TrainerEevee Trainer

7,72621338

$endgroup$

add a comment | 

3

$begingroup$

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 50 mins ago

J. W. TannerJ. W. Tanner

3,0701320

$endgroup$

• 
  
  
  

  
  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
  44 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
  30 mins ago
  

  
  
  
  

add a comment | 

Your Answer

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3

$begingroup$

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 56 mins ago

Eevee TrainerEevee Trainer

7,72621338

$endgroup$

add a comment | 

3

$begingroup$

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 56 mins ago

Eevee TrainerEevee Trainer

7,72621338

$endgroup$

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 56 mins ago

Eevee TrainerEevee Trainer

7,72621338

$endgroup$

share|cite|improve this answer

answered 56 mins ago

Eevee TrainerEevee Trainer

7,72621338

answered 56 mins ago

Eevee TrainerEevee Trainer

7,72621338

answered 56 mins ago

Eevee TrainerEevee Trainer

7,72621338

3

$begingroup$

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 50 mins ago

J. W. TannerJ. W. Tanner

3,0701320

$endgroup$

• 
  
  
  

  
  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
  44 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
  30 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

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 50 mins ago

J. W. TannerJ. W. Tanner

3,0701320

$endgroup$

• 
  
  
  

  
  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
  44 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
  30 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 50 mins ago

J. W. TannerJ. W. Tanner

3,0701320

$endgroup$

share|cite|improve this answer

answered 50 mins ago

J. W. TannerJ. W. Tanner

3,0701320

answered 50 mins ago

J. W. TannerJ. W. Tanner

3,0701320

answered 50 mins ago

J. W. TannerJ. W. Tanner

3,0701320