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How to modify this equation in order to find $x$?

1

$begingroup$

This comes from physics environment. Would anyone be able to help me alternate this equation in order to find the $x$?

I need to have this equation in the form of $x =
text{...something}$.
$$ A = frac{B}{C} - Bigg( frac{B}{C} - frac{B}{D} Bigg)e^{- frac{BF}{X}} $$
Please note: $e$ is an Euler's number here, and this is $e$ to the power of minus $frac{BF}{X}$ (looks little bit confusing).

My attempt:

$- frac{A - frac{B}{C}}{frac{B}{C} - frac{B}{D}} = e^{-frac{BF}{X}}$

$frac{D(AC-B)}{B(C-D)} = e^{-frac{BF}{X}}$

$ln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)} = -frac{BF}{X}$

$1 = -frac{BF}{Xln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)}}$

$X = -frac{BF}{ln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)}}$

However I'm not sure if it's correct.

Help's appreciated, thanks.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 24 at 12:16

weno

asked Mar 24 at 12:01

wenoweno

44311

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried to reorganize the terms isolating the exponential in one side of the equation and the rest on the other side?
  $endgroup$
  – Ertxiem
  Mar 24 at 12:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited my post with my attempted solution.
  $endgroup$
  – weno
  Mar 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you "lost" a minus sign between the 1st and the 2nd line of your computations.
  $endgroup$
  – Ertxiem
  Mar 24 at 12:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not really my computation, that part was done by wolfram.
  $endgroup$
  – weno
  Mar 24 at 12:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  My mistake, sorry. You solution seems correct to me.
  $endgroup$
  – Ertxiem
  Mar 24 at 12:24
  
  
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

This comes from physics environment. Would anyone be able to help me alternate this equation in order to find the $x$?

I need to have this equation in the form of $x =
text{...something}$.
$$ A = frac{B}{C} - Bigg( frac{B}{C} - frac{B}{D} Bigg)e^{- frac{BF}{X}} $$
Please note: $e$ is an Euler's number here, and this is $e$ to the power of minus $frac{BF}{X}$ (looks little bit confusing).

My attempt:

$- frac{A - frac{B}{C}}{frac{B}{C} - frac{B}{D}} = e^{-frac{BF}{X}}$

$frac{D(AC-B)}{B(C-D)} = e^{-frac{BF}{X}}$

$ln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)} = -frac{BF}{X}$

$1 = -frac{BF}{Xln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)}}$

$X = -frac{BF}{ln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)}}$

However I'm not sure if it's correct.

Help's appreciated, thanks.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 24 at 12:16

weno

asked Mar 24 at 12:01

wenoweno

44311

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Have you tried to reorganize the terms isolating the exponential in one side of the equation and the rest on the other side?
  $endgroup$
  – Ertxiem
  Mar 24 at 12:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've edited my post with my attempted solution.
  $endgroup$
  – weno
  Mar 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think you "lost" a minus sign between the 1st and the 2nd line of your computations.
  $endgroup$
  – Ertxiem
  Mar 24 at 12:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not really my computation, that part was done by wolfram.
  $endgroup$
  – weno
  Mar 24 at 12:21
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  My mistake, sorry. You solution seems correct to me.
  $endgroup$
  – Ertxiem
  Mar 24 at 12:24
  
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

This comes from physics environment. Would anyone be able to help me alternate this equation in order to find the $x$?

I need to have this equation in the form of $x =
text{...something}$.
$$ A = frac{B}{C} - Bigg( frac{B}{C} - frac{B}{D} Bigg)e^{- frac{BF}{X}} $$
Please note: $e$ is an Euler's number here, and this is $e$ to the power of minus $frac{BF}{X}$ (looks little bit confusing).

My attempt:

$- frac{A - frac{B}{C}}{frac{B}{C} - frac{B}{D}} = e^{-frac{BF}{X}}$

$frac{D(AC-B)}{B(C-D)} = e^{-frac{BF}{X}}$

$ln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)} = -frac{BF}{X}$

$1 = -frac{BF}{Xln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)}}$

$X = -frac{BF}{ln{Bigg(frac{D(AC-B)}{B(C-D)}Bigg)}}$

However I'm not sure if it's correct.

Help's appreciated, thanks.

algebra-precalculus logarithms

share|cite|improve this question

edited Mar 24 at 12:16

weno

asked Mar 24 at 12:01

wenoweno

44311

$endgroup$

share|cite|improve this question

edited Mar 24 at 12:16

weno

asked Mar 24 at 12:01

wenoweno

44311

share|cite|improve this question

edited Mar 24 at 12:16

weno

asked Mar 24 at 12:01

wenoweno

44311

asked Mar 24 at 12:01

wenoweno

44311

asked Mar 24 at 12:01

wenoweno

44311

1 Answer
1

active

oldest

votes

1

$begingroup$

Consider the problem of

$$A=G-He^{-frac{K}{X}}$$

We have

$$He^{-frac{K}{X}}=G-A$$

$$e^{-frac{K}{X}}=frac{G-A}{H}$$

Hence

$$-frac{K}{X}=logleft(frac{G-A}{H} right)$$

Hopefully you can continue from here.

Edit:

$$frac{K}{X}=logleft( frac{H}{G-A}right)$$

Hence

$$X=frac{K}{logleft( frac{H}{G-A}right)}$$

Now, $K=BF, G=frac{B}{C}, H=frac{B}{C}-frac{B}{D}$

Hence,

$$X=frac{BF}{logleft( frac{frac{B}{C}-frac{B}{D}}{frac{B}{C}-A}right)}=frac{BF}{logleft( frac{frac{BD-BC}{CD}}{frac{B-AC}{C}}right)}=frac{BF}{logleft( frac{B(D-C)}{D(B-AC)}right)}$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:13

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey, I edited my post with my solution. Would you be able to verify, please? Thanks.
  $endgroup$
  – weno
  Mar 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  our answers agree.
  $endgroup$
  – Siong Thye Goh
  Mar 24 at 12:25
  

  
  
  
  

add a comment | 

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1

$begingroup$

Consider the problem of

$$A=G-He^{-frac{K}{X}}$$

We have

$$He^{-frac{K}{X}}=G-A$$

$$e^{-frac{K}{X}}=frac{G-A}{H}$$

Hence

$$-frac{K}{X}=logleft(frac{G-A}{H} right)$$

Hopefully you can continue from here.

Edit:

$$frac{K}{X}=logleft( frac{H}{G-A}right)$$

Hence

$$X=frac{K}{logleft( frac{H}{G-A}right)}$$

Now, $K=BF, G=frac{B}{C}, H=frac{B}{C}-frac{B}{D}$

Hence,

$$X=frac{BF}{logleft( frac{frac{B}{C}-frac{B}{D}}{frac{B}{C}-A}right)}=frac{BF}{logleft( frac{frac{BD-BC}{CD}}{frac{B-AC}{C}}right)}=frac{BF}{logleft( frac{B(D-C)}{D(B-AC)}right)}$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:13

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey, I edited my post with my solution. Would you be able to verify, please? Thanks.
  $endgroup$
  – weno
  Mar 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  our answers agree.
  $endgroup$
  – Siong Thye Goh
  Mar 24 at 12:25
  

  
  
  
  

add a comment | 

1

$begingroup$

Consider the problem of

$$A=G-He^{-frac{K}{X}}$$

We have

$$He^{-frac{K}{X}}=G-A$$

$$e^{-frac{K}{X}}=frac{G-A}{H}$$

Hence

$$-frac{K}{X}=logleft(frac{G-A}{H} right)$$

Hopefully you can continue from here.

Edit:

$$frac{K}{X}=logleft( frac{H}{G-A}right)$$

Hence

$$X=frac{K}{logleft( frac{H}{G-A}right)}$$

Now, $K=BF, G=frac{B}{C}, H=frac{B}{C}-frac{B}{D}$

Hence,

$$X=frac{BF}{logleft( frac{frac{B}{C}-frac{B}{D}}{frac{B}{C}-A}right)}=frac{BF}{logleft( frac{frac{BD-BC}{CD}}{frac{B-AC}{C}}right)}=frac{BF}{logleft( frac{B(D-C)}{D(B-AC)}right)}$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:13

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hey, I edited my post with my solution. Would you be able to verify, please? Thanks.
  $endgroup$
  – weno
  Mar 24 at 12:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  our answers agree.
  $endgroup$
  – Siong Thye Goh
  Mar 24 at 12:25
  

  
  
  
  

add a comment | 

$begingroup$

Consider the problem of

$$A=G-He^{-frac{K}{X}}$$

We have

$$He^{-frac{K}{X}}=G-A$$

$$e^{-frac{K}{X}}=frac{G-A}{H}$$

Hence

$$-frac{K}{X}=logleft(frac{G-A}{H} right)$$

Hopefully you can continue from here.

Edit:

$$frac{K}{X}=logleft( frac{H}{G-A}right)$$

Hence

$$X=frac{K}{logleft( frac{H}{G-A}right)}$$

Now, $K=BF, G=frac{B}{C}, H=frac{B}{C}-frac{B}{D}$

Hence,

$$X=frac{BF}{logleft( frac{frac{B}{C}-frac{B}{D}}{frac{B}{C}-A}right)}=frac{BF}{logleft( frac{frac{BD-BC}{CD}}{frac{B-AC}{C}}right)}=frac{BF}{logleft( frac{B(D-C)}{D(B-AC)}right)}$$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:13

Siong Thye GohSiong Thye Goh

104k1468120

$endgroup$

share|cite|improve this answer

edited Mar 24 at 12:24

answered Mar 24 at 12:13

Siong Thye GohSiong Thye Goh

104k1468120

answered Mar 24 at 12:13

Siong Thye GohSiong Thye Goh

104k1468120

answered Mar 24 at 12:13

Siong Thye GohSiong Thye Goh

104k1468120