Having trouble with this log equation `

  • Thread starter Jeff Cook
  • Start date
  • Tags
    Log
In summary, the conversation is about solving for C in the equation log ( A - B - C ) = -C + A + B. After some initial confusion, it is determined that the original equation was incorrect and was later changed to A - log ( e^(A-B) / e^(C)) = A - B - C. One user suggests using the rule log(A) - log(B) = Log(A/B) to simplify the equation, while another user suggests using the Lambert's W function to solve it. After some back and forth, it is determined that the solution for C is C = A/2 - B. The conversation also touches on the topic of using different bases for logarithms.
  • #1
Jeff Cook
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
 
Physics news on Phys.org
  • #2
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
theperthvan said:
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).
 
  • #4
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
Jeff Cook said:
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.
 
  • #6
Jeff Cook said:
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.
 
  • #7
Got it...

Okay, thanks, guys.

J
 
  • #8
and remember that log(e^x) = x
 
  • #9
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...
 
  • #10
Gib Z said:
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.
 
  • #11
cristo said:
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.
 
  • #12
It looks like a tautology to me.
 
  • #13
tautology, how so?
 
  • #14
C=A/(ln10+1)-B

correct me if I'm wrong.
 
Last edited:
  • #15
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:
  • #16
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
uart said:
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.
 
  • #18
DanReit said:
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?
 
  • #19
Jeff Cook said:
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..
 
  • #20
Jeff Cook said:
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
Jeff Cook said:
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:
  • #21
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
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]
 
  • #23
DanReit said:
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:
 

1. What is a log equation?

A log equation is an equation that involves the logarithm function, which is used to solve for the unknown value in exponential equations. It is written in the form of logb(x) = y, where b is the base, x is the argument, and y is the result.

2. Why am I having trouble with this log equation?

Log equations can be challenging because they involve complex mathematical concepts and require a strong understanding of logarithms. It is important to review the rules and properties of logarithms and practice solving various log equations to improve your skills.

3. How do I solve a log equation?

To solve a log equation, you can use the inverse property of logarithms, which states that logb(x) = y is equivalent to by = x. This means you can rewrite the equation in exponential form and solve for the unknown value.

4. What are the common mistakes when solving log equations?

Some common mistakes when solving log equations include forgetting to apply the inverse property of logarithms, using the wrong base when rewriting the equation in exponential form, and misapplying the rules of logarithms. It is important to double-check your work and practice regularly to avoid these errors.

5. Are there any tips for solving log equations?

Yes, some helpful tips for solving log equations include simplifying the equation before solving, checking for extraneous solutions, and using a calculator for more complex calculations. It is also helpful to have a good understanding of algebraic manipulation and the properties of logarithms.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
10
Views
566
  • Precalculus Mathematics Homework Help
Replies
3
Views
729
  • Precalculus Mathematics Homework Help
Replies
9
Views
1K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
11
Views
4K
  • Precalculus Mathematics Homework Help
Replies
10
Views
2K
  • Precalculus Mathematics Homework Help
Replies
7
Views
2K
  • MATLAB, Maple, Mathematica, LaTeX
Replies
6
Views
2K
  • Precalculus Mathematics Homework Help
Replies
4
Views
1K
Replies
4
Views
695
Back
Top