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Help with $frac12 log_2 x - frac1log_2 x = frac76$

Need help with this linear equationStuck on an 'advanced logarithm problem': $2 log_2 x - log_2 (x - tfrac1 2) = log_3 3$Help with system of linear equationsI need help identifying the slope for an equation.Interval of the solutions to $log_1/2log_2(frac1+2x1+x)>0$ is?Solutions of $2^x 7^1/xle 14$Does $log_2 sqrt[4]4$ exist?Laws of logarithms: why isn't $frac14log_2(8x - 56)^16 - 12 = log_2((8x-56)^16)^frac 1 4 - 12$?How to find x if $log_2(x)cdotlog_2(x+2)=4$?Simplify $n^(log_2(n)/log_2(log_2(n)))$

3

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 Mar 11 at 5:56

user21820

39.4k543155

asked Mar 11 at 3:47

KevinKevin

496

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac 1 log x$ is not generally the same as $log frac 1 x $
  $endgroup$
  – J. W. Tanner
  Mar 11 at 3:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^-2/3$ ?
  $endgroup$
  – J. W. Tanner
  Mar 11 at 4:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
  $endgroup$
  – user21820
  Mar 11 at 5:45
  

  
  
  
  

add a comment | 

3

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 Mar 11 at 5:56

user21820

39.4k543155

asked Mar 11 at 3:47

KevinKevin

496

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $frac 1 log x$ is not generally the same as $log frac 1 x $
  $endgroup$
  – J. W. Tanner
  Mar 11 at 3:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Did you mean $x=mathbf 2^-2/3$ ?
  $endgroup$
  – J. W. Tanner
  Mar 11 at 4:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Welcome to Math SE, and thanks for providing your own thoughts on the problem. But next time use MathJax as required. It's not only much neater than handwriting, it's searchable.
  $endgroup$
  – user21820
  Mar 11 at 5:45
  

  
  
  
  

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 Mar 11 at 5:56

user21820

39.4k543155

asked Mar 11 at 3:47

KevinKevin

496

$endgroup$

share|cite|improve this question

edited Mar 11 at 5:56

user21820

39.4k543155

asked Mar 11 at 3:47

KevinKevin

496

share|cite|improve this question

edited Mar 11 at 5:56

user21820

39.4k543155

asked Mar 11 at 3:47

KevinKevin

496

edited Mar 11 at 5:56

user21820

39.4k543155

edited Mar 11 at 5:56

user21820

39.4k543155

asked Mar 11 at 3:47

KevinKevin

496

asked Mar 11 at 3:47

KevinKevin

496

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 Mar 11 at 3:50

Eevee TrainerEevee Trainer

7,87121339

$endgroup$

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 Mar 11 at 3:56

J. W. TannerJ. W. Tanner

3,2401320

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 11 at 4:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 11 at 4:16
  

  
  
  
  

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 Mar 11 at 3:50

Eevee TrainerEevee Trainer

7,87121339

$endgroup$

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 Mar 11 at 3:50

Eevee TrainerEevee Trainer

7,87121339

$endgroup$

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 Mar 11 at 3:50

Eevee TrainerEevee Trainer

7,87121339

$endgroup$

share|cite|improve this answer

answered Mar 11 at 3:50

Eevee TrainerEevee Trainer

7,87121339

answered Mar 11 at 3:50

Eevee TrainerEevee Trainer

7,87121339

answered Mar 11 at 3:50

Eevee TrainerEevee Trainer

7,87121339

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 Mar 11 at 3:56

J. W. TannerJ. W. Tanner

3,2401320

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 11 at 4:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 11 at 4:16
  

  
  
  
  

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 Mar 11 at 3:56

J. W. TannerJ. W. Tanner

3,2401320

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 11 at 4:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  Mar 11 at 4:16
  

  
  
  
  

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 Mar 11 at 3:56

J. W. TannerJ. W. Tanner

3,2401320

$endgroup$

share|cite|improve this answer

answered Mar 11 at 3:56

J. W. TannerJ. W. Tanner

3,2401320

answered Mar 11 at 3:56

J. W. TannerJ. W. Tanner

3,2401320

answered Mar 11 at 3:56

J. W. TannerJ. W. Tanner

3,2401320