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Homework Help: 2 Quick log problems

  1. Feb 8, 2010 #1
    I'm new here. You already knew that. Thank you, you, for clicking on my thread in any case. So, I've got these two log problems. My math teacher gave them out as a challenge. The class is relatively new to logs, and I'm having trouble working with these longer questions (longer than "simplify log5+log3"). I think my main issue is with the sub scripts. I don't really know what to do with them. For the first question, where it a normal algebra question, I'd have it no problem. Those logs get in the way though. We've been given the answers, but need to prove them. Any help would be appreciated.

    PS: the problems aren't for marks, they're more of a challenge thing before we do our test.

    1. The problem statement, all variables and given/known data

    1: log25+log2(x-3)=4


    3. The attempt at a solution

    log25+log2(x-3)=4
    log25(x-3)=4
    log2(5x-15)=4
    ???
    ???
    x= 31/5

    1. The problem statement, all variables and given/known data

    log3(x+4)=log77-log3(x+2)

    3. The attempt at a solution

    log3(x+4)=log77-log3(x+2)
    ???
    ???
    ???
    x= -9
    x=-5
     
  2. jcsd
  3. Feb 8, 2010 #2

    Mark44

    Staff: Mentor

    In your first problem, use the fact that log2a = b <===> a = 2b.
    In your second problem, transpose the log expression on the right side over to the left side, and use the fact that log77 = 1. Then you'll have a problem very similar to your first problem.
     
  4. Feb 8, 2010 #3

    Borek

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    Staff: Mentor

    Try to just apply definition of log.

    If

    [tex]log_ab = c[/tex]

    then

    [tex]b = a^c[/tex]

    Edit: Mark beat me.
     
  5. Feb 8, 2010 #4

    tiny-tim

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    Welcome to PF!

    Hi lexpar! Welcome to PF! :smile:
    If this was ln(5x-15) = 4, you'd just do eboth sides, wouldn't you?

    ok, so do 2both sides. :wink:
    Hint: log77 = … ? :smile:
     
  6. Feb 8, 2010 #5
    Thanks all three of you for the ultra speedy reply :D

    Figured out the first:

    log25+log2(x-3)=4
    log25(x-3)=4
    log2(5x-15)=4
    log25+log2x-log215=log216
    log25+log2x=log231
    5x=31
    x=31/5!!!! :D

    As for the second:

    log3(x+4)=log77-log3(x+2)
    log3((x+4)(x+2))= 1 (or maybe log33?)

    Where to go from here?
     
  7. Feb 8, 2010 #6

    tiny-tim

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    Same method as for the first one! :wink:
     
  8. Feb 8, 2010 #7

    Borek

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    Unfortunately - twice wrong.

    log(a-b) is NOT log(a) - log(b).
     
  9. Feb 8, 2010 #8

    In that case, what's the correct process? 31/5 is the correct answer, or so we were told.

    EDIT: Put a lot of thought into it. Read your earlier posts again:

    log2(5x-15)=4
    logab=c then b=ac
    therefore
    5x-15=24
    x=31/5


    Better? :D
     
    Last edited: Feb 8, 2010
  10. Feb 8, 2010 #9

    tiny-tim

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    Just write log2(5x-15) = log216 :smile:
     
  11. Feb 8, 2010 #10

    Borek

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    31/5 is a correct answer, but you got it by accident.

    Try 5x-15 = ... (just apply the definition, nothing more).
     
  12. Feb 8, 2010 #11

    Thanks, I got it :D Check out my last post, I edited it just for you <3
    *ahem*

    So still having a bit of trouble with the second question. I get to:

    log3((x+4)(x+2)=1
    log3(x2+6x+8)=1
    apply the definition=
    x2+6x+8 = 3

    But now I've got 2 x's! Where have I gone wrong??
     
  13. Feb 8, 2010 #12

    Borek

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    You listed two x's as correct answers. This is quadratic function. So everything seems to fit.

    Apart from the answers, seems to me like -9 is wrong. But I can be wrong, it is 1 a.m. here.
     
  14. Feb 8, 2010 #13

    Mark44

    Staff: Mentor

    So far, so good. Add -3 to both sides, and solve the quadratic equation. It's an easy one so you can factor it.

    Be sure to check both of your answers in the original equation, because one of them is extraneous.
     
  15. Feb 8, 2010 #14
    Thank you all for the responses. You guys really run a quality forum here :D

    Can I tempt you guys with one more problem? This one wasn't assigned, I'm just curious. I'm looking for a step by step walkthru here if you guys can give it, since this one flies right over my head.

    Question:

    logc20c/(logc50-logc5)

    Answer: log20c

    My progress:

    logc20c/(logc50-logc5)
    logc20c/(logc10)
    ...


    PS: sorry for the late response, was out playing pool with my dad. It's a very important school night activity :p
     
    Last edited: Feb 8, 2010
  16. Feb 8, 2010 #15

    Mark44

    Staff: Mentor

    This is fine, but put an = between the two expressions. The numerator could be written as logc(20) + 1, but I don't see that as an improvement.
     
  17. Feb 9, 2010 #16

    tiny-tim

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    Hi lexpar! :smile:

    (just got up :zzz: …)
    Always remember that logca = ln(a)/ln(c) = log(a)/log(c) = logba/logbc for any b. :wink:

    In this case, logc20c/(logc10) = … ? :smile:
     
  18. Feb 9, 2010 #17
    ...

    I'm really not catching on to that one :C
     
  19. Feb 9, 2010 #18

    tiny-tim

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    Write each of the top and the bottom of logc20c/(logc10) as fractions.
     
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