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Math Help - Few HW problems involving Logs, Natural logs, and Continuity.

  1. #1
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    Few HW problems involving Logs, Natural logs, and Continuity.

    Hi. I'm having problems with a few HW problems.

    1) log5 (x) + log5 (x+4) = ?

    How would I solve this? I get as far as log5 x^2+4x, but then I'm lost.

    2) (4x^3 - 5x^2 + 8x -7)/ ( x-1) for x<1
    -3x^2 +3x + a for x>=1

    solve for a.

    I can't seem to get rid of the (x-1) successfully

    3) e^(x+3) = e^(x) +7

    solve for x

    This one seems like it should be simple , but I can't solve it past

    x^2 +3x = ln7

    I don't really need the answers to these. I would just like to know how to solve them. I've been stuck on them for hours.

    Thanks
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  2. #2
    Super Member General's Avatar
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    (1)
    there is no equation!
    what is the right hand side?


    (2)
    Solve what?
    I do not see an equation. There is only a function!

    (3)
    Taking 'ln' of both sides will not solve it.
    Post your work.
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  3. #3
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    Quote Originally Posted by General View Post
    (1)
    there is no equation!
    what is the right hand side?


    (2)
    Solve what?
    I do not see an equation. There is only a function!

    (3)
    Taking 'ln' of both sides will not solve it.
    Post your work.
    Wow.. I completely screwed that up.... Sorry about that.

    Let's try that again:

    1) log5 (x) + log5 (x+4) = k
    solve for x in terms of k.

    2) (4x^3 - 5x^2 + 8x -7)/ ( x-1) for x<1
    -3x^2 +3x + a for x>=1

    What value for a will make this function continuous at 1?

    3) e^(x+3) = e^(x) +7

    For this I first brought the e's to one side so I got:

    e^(x+3) - e^(x) = 7

    Then I combined them using log(x) - log(y) = log(x/y). I think that's allowed in this case, is it not? After that I got:

    e^((x+3)/x) = 7

    Then I just used ln to get:
    (x+3)/x = ln7


    I must have done something wrong, I just don't know what.
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  4. #4
    Super Member General's Avatar
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    (1)
    You did the hard job :
    \log_5 (x^2+4x) = k

    We want to remove this noisy log. To do that, you need to apply this formula: a^{log_a{x}}=x. Got it?


    (2)
    Let f(x) be our function. f is continuous at x=1 \displaystyle \implies f(1) = \lim_{x \to 1^-} f(x)

    (3)
    Oh my god!
    No No No No No. \displaystyle e^a - e^b \neq e^{\frac{a}{b}}

    The property log(a)-log(b)=log(\frac{a}{b}) is valid for the logarithmic functions only, not for the exponential functions.

    To solve your eqaution, re-write it as : e^3 \cdot e^x - e^x = 7 . Can you complete it now?
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  5. #5
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    Okay. So for problem 1, I got that far, but it just looked funny and I couldn't figure out what to do. The 5^k is what really threw me off. I was trying to get a an integer answer I guess, and gave up. After you let me know I was on the right track, I got the right answer.

    2) I got that, and I get the 0/0 which of course means I need to do some work to get rid of the 0/0. However, I can't seem to figure out what to do. I can't get rid of the (x-1). I'm gonna keep working on it though.

    3) Lol... sorry for startling you. Now that I look at it, it makes a lot more sense to do what you did. However, sometimes it's just hard for me to see stuff, even though I know the rules, I just can't see that that rule should apply in this situation. At an rate, I got the answer for that one as well.

    Thanks for the help. Hopefully I get number 2 on my own, if not, I'll be back.

    Also, Is there a sticky for how to post like you did? like get the square root sign instead of typing sqrt, or showing an exponent instead of the carrot (^)? I'm sure I'll be back on these forums, and I think it's easier for everyone to view it like that.
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  6. #6
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    Quote Originally Posted by woody189 View Post
    [snip]
    Also, Is there a sticky for how to post like you did? like get the square root sign instead of typing sqrt, or showing an exponent instead of the carrot (^)? I'm sure I'll be back on these forums, and I think it's easier for everyone to view it like that.
    Click on the link in my signature.
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  7. #7
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    thanks
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  8. #8
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    For (1) : You will not get an integer answer. You will get an answer in terms of k.

    For (2) : Did you know how to deal with the 0/0 one?
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  9. #9
    Member rtblue's Avatar
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    for number one:

    \displaystyle \log_5 (x^2+4x) = k

    \displaystyle 5^k=x^2+4x

    \displaystyle x^2+4x-5^k=0

    Try using the quadratic formula from here.
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  10. #10
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    Sorry. I didn't check back to this site since I posted. I wound up getting the problems. Thanks for the help.
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