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Having trouble with this log equation `

  • Thread starter Jeff Cook
  • Start date
41
0
How would one go about solving for C in the following equation?

log ( A - B - C ) = -C + A + B

I have done this...

log ( A - B ) / log ( C ) = -C + A + B

Then...

C + ( log ( A - B ) / log ( C )) = A + B

But I am just shooting in the dark really. Where would one go from here? I would greatly appreciate any help with this.

Thanks,

Jeff
 
184
0
Step 1 is wrong.

log ( A - B - C ) doesn't equal log ( A - B ) / log ( C )

log ( A - B ) / log ( C ) is actually log [(A-B)/C]


And I don't think the rest can be done.
 
1,073
1
Step 1 is wrong.

log ( A - B - C ) doesn't equal log ( A - B ) / log ( C )

log ( A - B ) / log ( C ) is actually log [(A-B)/C]
That is wrong too. The rule is log(A) - log(B) = Log(A/B).
 
41
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I made an error, but I still have the question...

Okay, I made an error. I will step back to the original problem that I had but did not include.

In...

A - log ( e^(A-B) / e^(C)) = A - B - C

What is C? If it cannot be done, could speculate or explain why it can't?

Sorry about the confusion.

Again, any help with this would be greatly appreciated.

Thanks,

Jeff
 
1,073
1
Okay, I made an error. I will step back to the original problem that I had but did not include.

In...

A - log ( e^(A-B) / e^(C)) = A - B - C

What is C? If it cannot be done, could speculate or explain why it can't?

Sorry about the confusion.

Again, any help with this would be greatly appreciated.

Thanks,

Jeff
Use the rules of logs, starting with the one I posted above, this should simplify very easily.
 

cristo

Staff Emeritus
Science Advisor
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Okay, I made an error. I will step back to the original problem that I had but did not include.

In...

A - log ( e^(A-B) / e^(C)) = A - B - C

What is C? If it cannot be done, could speculate or explain why it can't?

Sorry about the confusion.

Again, any help with this would be greatly appreciated.

Thanks,

Jeff
Well, first subtract A from both sides. Next note that [tex]\frac{e^{(A-B)}}{e^{C}}=e^{(A-B)}e^{-C}=e^{(A-B-C)}[/tex] It should then simplify quite easily.
 
41
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Got it...

Okay, thanks, guys.

J
 
184
0
and remember that log(e^x) = x
 

Gib Z

Homework Helper
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Natural Log That is. I dont always like it when people use log for log base e, using ln is quicker and more informative...
 
184
0
Natural Log That is. I dont always like it when people use log for log base e, using ln is quicker and more informative...
Yes, same here.
 

uart

Science Advisor
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Well, first subtract A from both sides. Next note that [tex]\frac{e^{(A-B)}}{e^{C}}=e^{(A-B)}e^{-C}=e^{(A-B-C)}[/tex] It should then simplify quite easily.
If by simplified he means to express C in closed form in terms of A and B then I'd like to see someone do it. It looks like a transendental equation to me.
 

D H

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It looks like a tautology to me.
 

uart

Science Advisor
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tautology, how so?
 
5
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C=A/(ln10+1)-B

correct me if I'm wrong.
 
Last edited:

uart

Science Advisor
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correct me if I'm wrong.
You're wrong, just choose almost any two numbers for A and B and you'll find a counter-example. I tested your result with A=45 and B=3 and it definitely fails.
 
Last edited:

HallsofIvy

Science Advisor
Homework Helper
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Lambert's W function is defined as the inverse of the function f(x)= xex. It is, typically, the only way to solve an equation that has the unknown variable both as an exponent and not.
 
5
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You're wrong, just choose almost any two numbers for A and B and you'll find a counter-example. I tested your result with A=45 and B=3 and it definitely fails.

I'm sorry but your example is correct.

when A=45 and B=3 -> c=10.6256898

In the original equation the left part is 31.3747102; the right part is 42-10.6256898 -> which is exactly 31.3747102.
 

uart

Science Advisor
2,776
9
I'm sorry but your example is correct.

when A=45 and B=3 -> c=10.6256898

In the original equation the left part is 31.3747102; the right part is 42-10.6256898 -> which is exactly 31.3747102.
No, the original equation was log( A - B - C ) = -C + A + B and those numbers definitely do not work in that equation.

BTW, which equation are you substituting them into?
 
5
0
Okay, I made an error. I will step back to the original problem that I had but did not include.

In...

A - log ( e^(A-B) / e^(C)) = A - B - C

What is C? If it cannot be done, could speculate or explain why it can't?

Sorry about the confusion.

Again, any help with this would be greatly appreciated.

Thanks,

Jeff
Isn't that the original problem?

That is the equation I solved..
 

D H

Staff Emeritus
Science Advisor
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In...

A - log ( e^(A-B) / e^(C)) = A - B - C

What is C?
The above simplifies to
[tex]B+C = \log \left( \frac{e^{(A-B)}} {e^C}\right) = \log \left( {e^{(A-B-C)}}\right) = A-(B+C)[/tex]
Solving for C,
[tex]C=\frac A 2 - B[/tex]

Your original post is
How would one go about solving for C in the following equation?

log ( A - B - C ) = -C + A + B
This is an entirely different problem. So which statement of the problem is correct?
 
Last edited:
41
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There was an error in the first equation, so I backed up.

In...

A - log ( e^(A-B) / e^(C)) = A - B - C

What is C?

This is the issue at hand.

Jeff
 

D H

Staff Emeritus
Science Advisor
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Unless qualified with a base, 'log' usually means natural log. If that is what you meant by 'log', then [itex]C=\frac A 2 - B[/itex]. If you meant the base 10 logarithm instead, then [itex]C=\frac A {1+ln 10}-B[/itex]
 

uart

Science Advisor
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Isn't that the original problem?

That is the equation I solved..
Whoops, I didn't notice that the poster had changed the original problem as given in the first post.:blushing:
 

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