Calling math wizards (that means you!)

Working on some pre-calculus here, not too challenging for the math inclined.

http://img.photobucket.com/albums/v105/tman2093/DSC00615.jpg

I must be doing something wrong, because we haven't discussed imaginary numbers as a possible answer for a natural logarithm.

http://img.photobucket.com/albums/v105/tman2093/DSC00621.jpg

You can see how far I got, I eliminated the ln's, but I have that fraction to deal with now.

http://img.photobucket.com/albums/v105/tman2093/DSC00620.jpg

Calculator says this, not sure how it got there though.


Help please!

Where's Tangie?

The answer is 2.

:p

I could have answered this in 10 seconds 5 years ago. Today i haven't got a clue.



And yes, i do realise this reply is of absolutely no use to you :p

Brain.....too....rusty....with....logarithms. Sorry.

I could be wrong, but can't you take a number to a negative power in the numerator and put it into the denominator with a positive power? I seem to recall that's the easy way to get rid of a negative power.

So your e^-48 = -68 turns into:
1 / e^48 = -68

Brain.....too....rusty....with....logarithms. Sorry.

I could be wrong, but can't you take a number to a negative power in the numerator and put it into the denominator with a positive power? I seem to recall that's the easy way to get rid of a negative power.

So your e^-48 = -68 turns into:
1 / e^48 = -68


I'm gonna have to blame my handwriting on this. it's -4X, not -48

haha... there wouldnt have been anything to solve for if it was an 8... kinda worrying, fredjacksonsan :p

Okay, I'm coming up with the -4x = ln -68 as well, so you aren't alone. You can't take the natual log of a negative #. I must be stupid, on top of that... is it safe to assume that the second problem that you have entered on your calculator is a separate problem? Cause I have no idea how you got from the first problem to that problem.

tangie to the rescue

Tangie, yes, it's a seperate problem.

If I were to divide -4x = ln -68 by -1 to change the signs, It wouldn't affect the -68 would it? I believe I would get 4x = - ln -68, correct?

edit: 1500th post w00t.

Right, you'd just be changing the sign of the whole RS of the equation. Give me a second on that second problem, now that I know I'm not insane.

Okay, here's the deal... You can solve the second problem down to a cubic (use basic algebra skills). From there, I honestly just graphed it, and found the X intercept, which is of course, a solution to the cubic. The cubic you get from solving is not "factorable", so I chose to graph it to find the solution.

Tangie, Cubic equation is x^3 - 2x^2 - x -1, with solution of 2.54, which is correct. The reason that I didn't solve into a cubic equation sooner was because we've never discussed getting cubics from log problems. We've had quadratics, but not cubics.

Anyhow, thanks.

check the related links :thumbsup: lol

edit: d'oh! they used to be for a bunch of 4th grade math software

check the related links :thumbsup: lol

edit: d'oh! they used to be for a bunch of 4th grade math software
edHelper.com is a crappy site/link... and the 4th grade math "related links" have nothing to do with this... those are the only related links that come up for me.


Anyhoo, Isn't there some rule about taking the Ln of -e^any power? I should know this, but I need to spend some time grading papers instead of working on this. Sorry.

Oops. Sorry my previous answer was wrong. I forgot the last step. The answer is actually "Herman."

:lol2: :werd:

I hate maths.

Oops. Sorry my previous answer was wrong. I forgot the last step. The answer is actually "Herman."


You made a common mistake. You forgot to carry the 7, which altered the sequence. Follow order of operations, and you'll arrive at the correct answer of "platypus."

Anyway, I'll just put no solution for the negative natural log one.

Math = teh suxx0rz

Good luck :banghead:

Sorry. I should be grading papers now, as a matter of fact... I promised the kids they'd get them back tomorrow, since we had an "impromptu" fire drill today. Hope you get credit for those problems. I haven't done logarithms in YEARS (which tells you how important they are in math)...

Correct me if I'm wrong here, but how will this ever equal 75?

http://files.automotiveforums.com/gallery/watermark.php?file=/503/63345Graph1.jpg

Has this forum go so god damn patheticly boring that the most interesting thread in it is about solving equations?

slutty- You're right, from the graph the only possible solution is 0(or close to it), but it makes a false statement. Therefore, no solution.

Moppie- Funny, ain't it?

Would you verify that the equation is correct? I have down: 7-e^(-4x)=75 as your equation.

For this statement to be true, the term -e^(-4x) must equate to a negative number as you cannot take 7 from any positive number and get 75.

However, if you graph e^x, you will see that it is asymtopic to zero and thus never become negative. Therefore, I would deduce that there is no answer to this probem.

Maybe it was 7+e^(-4x)=75 and not 7-e^(-4x)=75?

I hope I make sense here :icon16:

a.

Bored at work so I decided to take another look... I made a couple graphs on Excel to illustrate my point...

http://files.automotiveforums.com/gallery/watermark.php?file=/503/39429lnx.jpg

http://files.automotiveforums.com/gallery/watermark.php?file=/503/39429ex.jpg

So there we go... the function e^(-4x) cannot go negative thus this, at least in my deduction, leads me to believe that the LHS of the equation will never be greater than or equal to 7.

a.

Good looking out, but I got my test back today.

I aced it, so no solution HAS to be right for 7-e^(-4X)=75. I checked before completing the test(we took it over the course of 2 days) and I copied the problem correctly. If I made 100, then the answer must be no solution

Thanks to all that have offered help!