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Logarithmic problem

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data

    4log45

    2. Relevant equations

    ay = x ~~~> loga(x) = y

    3. The attempt at a solution

    I've separated the problem and decided to solve the logarithmic section first giving me:

    4x = 5

    I could move forward if I wasn't stumped on how to figure out what x is in this equation. This is probably very simple and maybe my mind isn't comprehending well this morning, but I've stared at this problem for quite some time now. I appreciate your assistance in advance.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 21, 2012 #2

    jedishrfu

    Staff: Mentor

    given a number x, what is does its log represent?

    knowing that you should be able to solve this immediately in your head.
     
  4. Mar 21, 2012 #3
    I'm sorry, but I'm not understanding what your trying to say. Would you mind clarifying that for me.
     
  5. Mar 21, 2012 #4

    jedishrfu

    Staff: Mentor

    have you tried it on a calculator yet to see what the answer is?
     
  6. Mar 21, 2012 #5
    I would type it in and work backwards, but my ti84 automatically sets the log with a base of 10 and I don't know how to change it to a base of 4. Do you know how to do this?

    When I get stumped, I usually type it into the calculator and work backwards to figure it out, but in this case I can't.

    How would I type this into a TI84?
     
  7. Mar 21, 2012 #6

    jedishrfu

    Staff: Mentor

    ok try it with base ten and see what you get.

    10^log(5)
     
  8. Mar 21, 2012 #7
    This is what I typed in:

    4(log(5)) and this is what I got:

    2.795880017

    Now what?
     
  9. Mar 21, 2012 #8

    jedishrfu

    Staff: Mentor

    you're missing the point, did you try 10^log(5)?
     
  10. Mar 21, 2012 #9

    jedishrfu

    Staff: Mentor

    Going back to your original work:

    4x = 5

    take the log base 4 of both sides:

    x * log44 = x = log45
     
  11. Mar 21, 2012 #10
    :biggrin:

    Gotcha! I see what your saying now. If my subscript is 4 and my calculator has a default of 10, all I need to do is sub out 10 when I see 4 and I will get the same answer.

    Is this correct?

    For example:

    4log45 would give me the same answer if I used 10log105

    I've typed it in and got the right answer.

    Thank you so much for your help, I really appreciate it. I promise I'm usually good at math, this one just stumped me.
     
  12. Mar 21, 2012 #11

    jedishrfu

    Staff: Mentor

    what you have is a kind of identity. This doesn't work in the general case as you were thinking. The teacher selected this example because you can't just enter it into the calculator to get the answer.

    The identity is h/she wanted you to notice is:

    base^logbase(x) = x

    so it could be 10^log10(5) = 5 or e^ln(5) = 5 ...

    a more complicated example might be:

    8^log4(5) = ??? with this variation you'd need to understand how to convert between bases
     
  13. Mar 21, 2012 #12
    Ok, so in my case with my bases being equal, my answer will always be x.

    We haven't gone over this yet, but how would I convert between bases in the more complicated problem of 8^log4(5)?

    Would you mind working this problem out, showing the steps so I can see?
     
  14. Mar 21, 2012 #13

    jedishrfu

    Staff: Mentor

    I chose poorly lets try 16^log4(5) = x

    taking the log of both sides yields:

    log4(5) = log16(x)

    next applying the log conversion:

    loga(x) = logb(x) / logb(a)

    log16(x) = log4(x) / log4(16)

    and since 16 is 4^2 then log4(16) = 2

    log4(5) = log16(x) = log4(x) / log4(16) = log4(x) / 2

    so we now have:

    2 * log4(5) = log4(x)

    and bringing the 2 inside

    log4(5^2) = log4(x)

    and lastly 5^2 = 25 and hence x=25

    my original 8^log4(5) would have yielded x = 5 * sqrt(5) or 5^(3/2) since the
    conversion factor log4(8) = log4(4*2)=log4(4)+log4(2) = 1 + 1/2 = 3/2
     
  15. Mar 21, 2012 #14
    Thank you, now I have a head start for next weeks lesson. I really appreciate your help with this.
     
  16. Mar 21, 2012 #15

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Basically, the first thing you should have learned about "[itex]log_a(x)[/itex]" is that it is the inverse function to [itex]f(x)= a^x[/itex]. That is, if [itex]y= a^x[/itex] then [itex]x= ln_a(y)[/itex]. Putting those together, [itex]ln_a(y)= ln_a(a^x)= x[/itex] and [itex]a^{x}= a^{ln_a(y)}= y[/itex].
     
    Last edited: Mar 23, 2012
  17. Mar 22, 2012 #16
    Thanks for all the help you guys have provided me with. I really appreciate it. This class I'm in right now is rough because the teacher isn't explaining anything clearly. She just starts talking as if everyone knows what she's talking about and gives no explanation as to why were doing things. I'm one of those people who needs at least a little bit of a "why" to what I'm doing. I want to know where it's leading.

    Anyways sorry for my rant. Thanks again.
     
  18. Mar 22, 2012 #17

    jedishrfu

    Staff: Mentor

  19. Mar 23, 2012 #18
    [itex]x=4^{log_45}[/itex]
    Taking [itex]log_4 [/itex] both side
    Whatever you do on the left of the equation, you should do the same on the right.
    It doesn't matter if you add, multiply, square or any other mathematical operations.

    [itex]log_4x=log_44^{log_45}[/itex]
    [itex]log_4x=log_45log_44[/itex] You should remember this rule. lgx^y=ylgx
    [itex]log_4x=log_45[/itex]

    you can find the value of x
     
  20. Mar 23, 2012 #19
    Thanks for the Khan Academy web address, it's really simple to understand when it's explained clearly and things aren't left out. And thank you azizlwl for that great break down.
     
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