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!

Log Simplification Function

  1. Jan 18, 2012 #1

    MacLaddy

    User Avatar
    Gold Member

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

    Assume [itex]log_bx=0.36, log_by=0.56, log_bz=0.83[/itex]

    Evaluate the following expressions

    [tex]log_b\frac{\sqrt{xy}}{z}[/tex]

    2. Relevant equations



    3. The attempt at a solution

    I know that the first step of simplification is [itex](log_b\sqrt{x}+log_b\sqrt{y})-(log_bz)[/itex], and that [itex](log_bz) = 0.83[/itex], so that ends up being [itex](log_b\sqrt{x}+log_b\sqrt{y})-(0.83)[/itex]

    However, I have no idea what to do with the [itex]\sqrt{x}[/itex] that is in the first two logs. I can't seem to find the right search online for that, and it's not in my book.

    Any help would be greatly appreciated.

    Thanks,
    Mac
     
  2. jcsd
  3. Jan 18, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    log(sqrt(x))=log(x^(1/2))=(1/2)*log(x). Does that help?
     
  4. Jan 18, 2012 #3

    MacLaddy

    User Avatar
    Gold Member

    I guess I should clarify what I am asking.

    If [itex]log_bx = 0.36,[/itex] then does [itex]log_b\sqrt{x}=\sqrt{0.36}[/itex]
     
  5. Jan 18, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Nooo. [itex]log_b\sqrt{x}=log_b x^\frac{1}{2}=\frac{1}{2}*log_b{x}[/itex]. Does TeXing it make it clearer?
     
  6. Jan 18, 2012 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    The quick answer ... No.

    By the rule of logarithms. [itex]\displaystyle\log_b(\sqrt{x})=\frac{1}{2}\log_b(x)\,.[/itex]

    For one thing, [itex]\sqrt{x}=x^{1/2}\,.[/itex]
     
  7. Jan 18, 2012 #6

    MacLaddy

    User Avatar
    Gold Member

    Ahh, apparently TeXing does make it clearer. I guess I didn't read your first answer correctly.

    I had forgotten about that Log rule, but I'm still a bit rusty. Do all exponents in a logarithm always get multiplied at the front? It seems that they do.

    I appreciate your help.
     
  8. Jan 18, 2012 #7

    MacLaddy

    User Avatar
    Gold Member

    Thanks Dick and SammyS. That is exactly what I was looking for.
     
  9. Jan 18, 2012 #8

    Mentallic

    User Avatar
    Homework Helper

    Yes, in general

    [tex]\log(a^b)=b\cdot\log(a), a>0[/tex]

    If we look at some special cases such as b=0, this gives us [tex]\log(a^0)=0\cdot\log(a)=0[/tex] which is true because of the log of 1 is always 0.

    b=-1 gives

    [tex]\log(a^{-1})=-\log(a)[/tex] which can also be seen in another way as
    [tex]\log(a^{-1})=\log(\frac{1}{a})=\log(1)-\log(a)=0-\log(a)[/tex]
     
  10. Jan 18, 2012 #9

    MacLaddy

    User Avatar
    Gold Member

    Thanks Mentallic. I think I'll print that for my notes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Log Simplification Function
Loading...