1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I know the very basic principles of logs

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    I'm new to logs. I know the very basic principles of logs, unfortunately my notes are not helping my on this occasion. What I have so far is Solve the value 'x'


    2. Relevant equations

    6^(x+3)=56.34

    3. The attempt at a solution
    Here's what I have done, and I'm not sure where to go from here so could someone explain?

    log6^(x+3)=log5.34
    (x+3)log6=log56.34
    (x+3)log6=1.751
    ------

    Not sure if I've done it correctly, and I'm not sure where to go from next :/

    If you have seen me asking a lot of questions tonight, I have an exam next week and I'm just getting ready for it :)
     
  2. jcsd
  3. Dec 3, 2011 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Re: Logs/Logarithm

    Good. Now you have an equation of the form

    [tex](x+3)b=c[/tex]

    Solve for x.
     
  4. Dec 3, 2011 #3
    Re: Logs/Logarithm

    So, the next line would be xlog6 +3log6 = 1.751

    Yes or no?
     
  5. Dec 3, 2011 #4

    BruceW

    User Avatar
    Homework Helper

    Re: Logs/Logarithm

    I'm guessing on your first line, you meant log56.34 (since that's what you put on the next line, which is correct).

    You have got it right so far. From here, its simple rearrangement to get x.

    Edit: Sorry, I started writing this before micromass posted, so I am a little behind here.

    yes, that would be the next line (although you could instead have divided both sides by b), but you get the right answer either way.
     
  6. Dec 3, 2011 #5

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Re: Logs/Logarithm

    Yes.
     
  7. Dec 3, 2011 #6

    Mark44

    Staff: Mentor

    Re: Logs/Logarithm

    As already noted, the first line above has a typo.
    As an alternative, starting from (x + 3)log(6) = log(56.34),
    you could divide both sides by log(6) to get x + 3 = log(56.34)/log(6), and you can solve for x from there.

    Note that put back in log(56.34) instead of using your value of 1.751. You'll get a more precise result if you DON'T put in rounded intermediate values. If you use the full values that your calculator gives you, you'll get answers that don't have built-in inaccuracies.
     
  8. Dec 3, 2011 #7
    Re: Logs/Logarithm

    So, on that last line I done... xlog6 + 3log6

    Do I (on a calculator) type log6= then Ans*3... and then add it to log6 ? I'm confused now :(
     
  9. Dec 3, 2011 #8
    Re: Logs/Logarithm

    Any help guys? I'm real stuck ! Can't move on from here...
     
  10. Dec 3, 2011 #9
    Re: Logs/Logarithm

    So I couldn't do multiplication method so I went for division.. Here's what I got.

    (x+3=1.751)/log6

    Uhh doesn't look right but is it ?
     
  11. Dec 3, 2011 #10

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Re: Logs/Logarithm

    I really don't know what is bothering you so much. Can you solve

    [tex]2(x+3)=5[/tex]

    ??

    The equation you list is the same thing but with different numbers.
     
  12. Dec 3, 2011 #11
    Re: Logs/Logarithm

    So I think I have a final answer, I kept simplifying down. after that previous line i got (x=1.751)/log6 (-3)

    Which then gave me a final answer of -0.75 to 2dp
     
  13. Dec 3, 2011 #12
    Re: Logs/Logarithm

    New Question!!!

    So I got another Question because I didn't full click with that last one..

    75.2^x = 10.5 ^(3x-5)


    What I got so far...

    log75.2^x=log10.5^(3x-5)
    xlog75.2 = (3x-5)log10.5

    Trying the next line... Suppose its the same as last time but the other way around?

    Edit:
    [tex](xlog75.2=3x-5)/log10.5[/tex]
     
  14. Dec 3, 2011 #13

    BruceW

    User Avatar
    Homework Helper

    Re: Logs/Logarithm

    You got the last question right, with -0.75.

    For the new question, you have got the use of logarithm correct. Now it is just a case of rearranging.
     
  15. Dec 3, 2011 #14
    Re: Logs/Logarithm

    The new question I re-wrote the equation like so, [tex] xlog75.2 = 3xlog10.5-5log10.5[/tex] Is that wrong and/or a better method than the last way?
     
  16. Dec 3, 2011 #15

    Mark44

    Staff: Mentor

    Re: Logs/Logarithm

    You're not done yet. Add -3xlog(10.5) to both sides. Then on the left side, you'll have two terms with a factor of x. Factor the two terms into x(...other stuff) = <stuff on the right side>.
     
  17. Dec 3, 2011 #16

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Logs/Logarithm

    You cannot put equals signs inside of expressions within parentheses. It makes no sense.
     
  18. Dec 3, 2011 #17
    Re: Logs/Logarithm

    Sorry, not quite sure what you mean here?

    Thanks for your help Mark, but I'm still unsure now :( Would it look like this..
    [tex](-3xlog(10.5)xlog75.2=-3xlog(10.5) 3x-5) / log10.5 [/tex]

    Or [tex]((-3xlog(10.5)xlog75.2=(3xlog(10.5))3xlog10.5-5log10.5)[/tex] CONFUSED ! :(
     
  19. Dec 3, 2011 #18

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: Logs/Logarithm

    I was talking about this:

    You have an equals sign inside the parentheses. It makes no sense.
     
  20. Dec 3, 2011 #19

    BruceW

    User Avatar
    Homework Helper

    Re: Logs/Logarithm

    haha, yeah, I think he means 'divide both sides by log10.5'

    Edit: I don't know if that is allowable notation in maths, but you can see what he means by it.
     
  21. Dec 3, 2011 #20

    BruceW

    User Avatar
    Homework Helper

    Re: Logs/Logarithm

    Think it through yourself, you've done the hard bit already, you just needed to rearrange to get x as the subject of the formula.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: I know the very basic principles of logs
  1. Basic LOG problem (Replies: 2)

Loading...