I need some math help

Discussion in 'Locker Room' started by Sharpy aint SAWFT, Apr 8, 2014.

  1. I attached the picture of the math problem, it has to do with logarithms and such and I'm drawing a blank at the moment, I got like an hour before class and was seeing if anyone could offer some help.

    Attached Files:

  2. Shit, I knew how to do this stuff. I'll get back at you but probably not. Maybe @Delik
  3. That's where I'm at as well, I've been at this for like 2 hours (the assignment not just the problem) this is just the last one and I want to get 100% cuz it's a take home exam haha. I have this same thread up on a few other sites trying to get an answer, can't believe the brain fart im having.. My last math class before getting my 2 yr university degree.
  4. Yeah, dude. I totally forgot this stuff.
    I'm thinking of two different subjects while trying to do this. Like, I'm bringing the e4X from the right side to the left and start doing division in the bottom and my other my guess was e12x then you bring over e4x. Confusing stuff and I should probably start learning this stuff again. Yeah, I'm lost right now. Sorry dude.

    You should run around campus and start asking people for help. lol
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  5. Lol I started like this and tis where I got the brain fart, I'm bout to just ask someone before class pfft, hopefully someone got it right.
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  6. e3x - e4x = e4x - 15

    e-x = e4x - 15

    -3 = -3 * 4 - 15

    x = -3
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  7. Thanks for this, it's the best response I've received and it makes sense to me in a my mind is too burnt out and I can see this being a good answer, at least one that will get me partial credit if wrong for some chance.
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  8. I'm guessing * = multiply

    -3 = -3 x 4 - 15
    -3 = -12 - 15
    -3 = -27

    Kind of confusing, can you clear this up?
  9. I have no idea, you're right though that does give -27. These always confused me :dawg:
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  10. Wait, if these are supposed to involve logarithms, isn't your math completely incorrect.
  11. Probably. I'm ill, tired and bored. :dawg:
  12. Yep, you're going crazy Jono, get to bed.
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  13. Those look like powers of to me. As in e^(3x) * (e^4)^x if memory serves right you would add the powers of together when multiplying (?) so it would be e^7x I think

    EDIT: But then the e's cancel since you are dealing with powers of and you essentially come out with 7x = 4x-15. And thats tit from there. But I don't feel like thats right.
  14. It's logarithms.
  15. (e^3x) - (e^4)x = e^4x - 15

    e^7x = e^4x - 15

    e^3x = -15

    x = -5

    I think this should be right now.
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  16. ^See don't listen to me
  17. I'm pretty sure it's logs, not powers though. Something to do with the e's.
  18. alright so here's how its done, finally figured it out..

    e^3x times e^4x = e^4x-15 basically you just cancel out the e^4x on both sides and get e^3x = e^-15 and then from there you make it 3x = -15 and get x = -5 if you solve for x. it was a stupid problem.
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  19. So @Delik had it right last time he posted I think. I put him on the right path though! :dawg:
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  20. If it's exponential I'd not believe that's how it works..
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