Having trouble with this log equation `

[Jeff Cook]

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

 

[theperthvan]

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.

 

[d_leet]

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).

 

[Jeff Cook]

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

 

[d_leet]

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]

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 $$\frac{e^{(A-B)}}{e^{C}}=e^{(A-B)}e^{-C}=e^{(A-B-C)}$$ It should then simplify quite easily.

 

[Jeff Cook]

Got it...

Okay, thanks, guys.

J

 

[theperthvan]

and remember that log(e^x) = x

 

[Gib Z]

Natural Log That is. I don't always like it when people use log for log base e, using ln is quicker and more informative...

 

[theperthvan]

Natural Log That is. I don't always like it when people use log for log base e, using ln is quicker and more informative...

Yes, same here.

 

[uart]

Well, first subtract A from both sides. Next note that $$\frac{e^{(A-B)}}{e^{C}}=e^{(A-B)}e^{-C}=e^{(A-B-C)}$$ 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]

It looks like a tautology to me.

 

[uart]

tautology, how so?

 

[DanReit]

C=A/(ln10+1)-B

correct me if I'm wrong.

 

[uart]

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.

 

[HallsofIvy]

Lambert's W function is defined as the inverse of the function f(x)= xe^x. It is, typically, the only way to solve an equation that has the unknown variable both as an exponent and not.

 

[DanReit]

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]

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?

 

[DanReit]

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]

In...

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

What is C?

The above simplifies to
$$B+C = \log \left( \frac{e^{(A-B)}} {e^C}\right) = \log \left( {e^{(A-B-C)}}\right) = A-(B+C)$$
Solving for C,
$$C=\frac A 2 - B$$

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?

 

[Jeff Cook]

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]

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

 

[uart]

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.