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Logorithm equations help

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data
    How can I solve these? Only using logarithms and(or) exponential(not natural logarithms, or e).

    34x-1=7

    35x/76x=4

    3. The attempt at a solution

    I have tried turning the first one into log10. By doing Log34x-1=7 to Log104x-1/Log103=Log107, then somehow turning that into Log104x+Log10-1/Log103=Log107. But that doesn't seem right. It's where I'm stuck right now, can anyone help? Thank you.
     
    Last edited: Feb 20, 2013
  2. jcsd
  3. Feb 20, 2013 #2

    SammyS

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    [itex]\displaystyle \log_{3\,}(4x-1)=7\ \ [/itex] is equivalent to [itex]\displaystyle \ 3^{7}= 4x-1\,, \ [/itex] not the given equation.
     
  4. Feb 20, 2013 #3
    Take the logarithm of both sides (just the usual base 10). What's the rule about exponents -- for say "b" in log(a^b)?
     
  5. Feb 20, 2013 #4
    I just realized that. I meant to say Log37=4x-1
     
  6. Feb 20, 2013 #5
    Such as a*log(b)? He corrected me. But I'm trying to see if that could help.
     
  7. Feb 20, 2013 #6

    SammyS

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    Well, solve this for x.
     
  8. Feb 20, 2013 #7
    Thank you, it is, solving another equation like it I got the answer. I had my suspensions but due to having mismarked things I got I couldn't go through. Now that second one's where I'm stuck. It seems that I'd have to do Log5/64=5x/6x or something that wacky. But can you or someone else help me? Thank you.
     
  9. Feb 20, 2013 #8
    I did Log10(7)/Log10(3)=4x-1. I got a decimal number(when pluged into calculator), then solved it for X with what I got. Is that right?
     
  10. Feb 20, 2013 #9

    SammyS

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    I would be inclined to write the original equation as [itex]\displaystyle \
    \left(\frac{3^5}{7^6}\right)^{x}=4\ .[/itex]

    Then take both sides to the 1/x power, ...
     
  11. Feb 20, 2013 #10
    If both sides are equal, then X is correct. My approach would be be to log3 both sides to bring down the (4x-1)
     
  12. Feb 20, 2013 #11
    I'm new to Logs, we just had the chapter this week, but you mean log the other side of the equal sign as well?
     
  13. Feb 20, 2013 #12
    Like 41/x while canceling the other side? Sorry for sounding like someone with no clue of anything. But this chapter confuses.
     
  14. Feb 20, 2013 #13

    SammyS

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    [itex]\displaystyle \left(a^x\right)^{1/x}=a[/itex]

    Yes, if you mean [itex]\displaystyle \ \frac{3^5}{7^6}=4^{1/x}\ .[/itex]

    Now write the logarithmic version of this equation.
     
  15. Feb 21, 2013 #14
    So Log435/76=1/x ? Then solve using log10 right?
     
  16. Feb 21, 2013 #15

    NascentOxygen

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    Why not solve it the way you think is probably right, then substitute back your answer for x to see whether it fits the original equation?

    That way, you'll quickly see whether you are on the right track.

    You might even have a calculator which does log4. Otherwise, yes, use log10 and do the necessary conversion.
     
  17. Feb 21, 2013 #16
    I did that. I don't have my paper with me but I think that I had a negative -0.4(or this might have been for another question). Substituting it back in I got a decimal very close(.999something) to the answer. Thank you all very much for everything. I would have preferred something like a fraction but this is better than nothing. Thank you again.
     
  18. Feb 21, 2013 #17

    Ray Vickson

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    Your equations are certainly wrong as you have written them. You write Log34x-1, which means ##\log_{3}(4x) - 1##. Maybe you really mean Log3(4x-1)? If so, that is what you need to write!
     
  19. Feb 21, 2013 #18

    SammyS

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    Yes.

    You can use some properties of log to simplify this.

    [itex]\displaystyle \frac{1}{x}=5\log_4(3)-6\log_4(7)[/itex]

    Use the change of base formula if you need to use log10.
     
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