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Math Help - Help With Log Notation

  1. #1
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    Help With Log Notation

    What would ln stand for in a log equation?

    For example 8ln x = 16 ? Is that the same as 8 log x = 16 ?

    Thanks!
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  2. #2
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    \ln(x) = \log_e(x)
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  3. #3
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    Thanks for the quick response! Does e = 10 is the base if it not stated, just like normal logs?
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  4. #4
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    Quote Originally Posted by mathguy20 View Post
    Thanks for the quick response! Does e = 10 is the base if it not stated, just like normal logs?
    e \neq 10 under any circumstance. In general however: \ln(x) = \log_e(x) and \log(x) = \log_{10}(x) where the base isn't explicitly stated
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  5. #5
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    Got it -- here's another log question if you don't mind.

    How would I solve for x?

    2log_4(x) - log_4(x+3) = 1
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  6. #6
    Super Member Quacky's Avatar
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    We have: Log_4{(x^2)}-Log_4{(x+3)}=1

    Can you see why?

    So combine logs using the Log(a)-Log(b)=Log(\frac{a}{b}) and see if you get anywhere.
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  7. #7
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    Yeah - so would it be Log_4(\frac {x^2}{x+3}) = 1 ?

    I think I got lost in the base 4, because I've only done those in base 10.

    Then, I can just solve for x from 4 = \frac{x^2}{x+3}. So x = {-2, 6} ?
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  8. #8
    Super Member Quacky's Avatar
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    Quote Originally Posted by mathguy20 View Post
    Yeah - so would it be Log_4(\frac {x^2}{x+3}) = 1 ?

    I think I got lost in the base 4, because I've only done those in base 10.

    Then, I can just solve for x from 4 = \frac{x^2}{x+3}. So x = {-2, 6} ?
    Yeah. Remember that you can always try it and substitute your answer into the original equation to check!

    Edit: although x can't be -2 because this isn't defined by your original equation. Be careful with logs; this happens frequently!
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  9. #9
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    Quote Originally Posted by Quacky View Post
    Yeah. Remember that you can always try it and substitute your answer into the original equation to check!
    I see how I can plug in -2 and 6 to the equation we got in the final steps, but is there a way I can plug that into the original equation and check it on my TI-84? (I'm not sure how to use bases other than 10 on the calculator.)
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  10. #10
    Super Member Quacky's Avatar
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    I am unfamiliar with this exact calculator, but I would expect, if there isn't a button that calculates logs to all bases, that you can, using the change of base rule. If you convert the logs from base 4 to base 10, then you can plug the numbers in. I don't know whether there's a shorthand, my calculator can calculate logs to all bases for me so I don't have to worry.
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  11. #11
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    Hello mathguy20,
    log B4 (4) =1. Bring all terms to LHS.Combine all the log terms to 1 term= to 0.Solve for x



    bjh
    Last edited by bjhopper; May 2nd 2011 at 07:13 PM. Reason: misspelling
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