# Homework Help: Having trouble with this log equation `

1. Jan 1, 2007

### 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

2. Jan 1, 2007

### 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.

3. Jan 1, 2007

### d_leet

That is wrong too. The rule is log(A) - log(B) = Log(A/B).

4. Jan 1, 2007

### 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

5. Jan 1, 2007

### d_leet

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

6. Jan 1, 2007

### cristo

Staff Emeritus
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.

7. Jan 1, 2007

### Jeff Cook

Got it...

Okay, thanks, guys.

J

8. Jan 1, 2007

### theperthvan

and remember that log(e^x) = x

9. Jan 3, 2007

### Gib Z

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

10. Jan 3, 2007

### theperthvan

Yes, same here.

11. Jan 4, 2007

### uart

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.

12. Jan 4, 2007

### D H

Staff Emeritus
It looks like a tautology to me.

13. Jan 4, 2007

### uart

tautology, how so?

14. Jan 4, 2007

### DanReit

C=A/(ln10+1)-B

correct me if I'm wrong.

Last edited: Jan 4, 2007
15. Jan 5, 2007

### uart

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: Jan 5, 2007
16. Jan 5, 2007

### HallsofIvy

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.

17. Jan 5, 2007

### DanReit

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.

18. Jan 5, 2007

### uart

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?

19. Jan 5, 2007

### DanReit

Isn't that the original problem?

That is the equation I solved..

20. Jan 5, 2007

### D H

Staff Emeritus
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
This is an entirely different problem. So which statement of the problem is correct?

Last edited: Jan 5, 2007
21. Jan 5, 2007

### 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

22. Jan 5, 2007

### 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 $C=\frac A {1+ln 10}-B$

23. Jan 5, 2007

### uart

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

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