# 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

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#### 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?

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?

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
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?

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

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

#### theperthvan

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

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

Staff Emeritus
It looks like a tautology to me.

#### uart

tautology, how so?

#### DanReit

C=A/(ln10+1)-B

correct me if I'm wrong.

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#### 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.

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#### HallsofIvy

Homework Helper
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.

#### 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?

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
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,
$$C=\frac A 2 - B$$

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?

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#### 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

Staff Emeritus
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

Whoops, I didn't notice that the poster had changed the original problem as given in the first post. 