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Using Properties of Logarithms

  1. Oct 13, 2016 #1
    1. The problem statement, all variables and given/known data

    Question: Use the properties of logarithms to rewrite and simplify the logarithmic expression.

    1. [tex]log_{5}\dfrac{1}{250}[/tex]

    2. Relevant equations
    Product Property: [tex]log_{a}(uv) = log_{a}u + log_{a}v[/tex]
    Quotient Property: [tex]log_{a}\dfrac{u}{v} = log_{a}u - log_{a}v[/tex]
    Power Property: [tex]log_{a}u^{k} = k\cdot log_{a}u[/tex]

    3. The attempt at a solution

    [tex]log_{5}1 - log_{5}250[/tex]

    I know when it's dividing, you split it up by subtracting. (Still not confident in the first step) Do I find out 5 to what power gives it 250? Even if I did that, it won't match 250. Pretty sure I'm missing something I'm unaware of.

    Answer: [tex]-3 - log_{5}2[/tex]
     

    Attached Files:

  2. jcsd
  3. Oct 13, 2016 #2

    SammyS

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    What is the prime factorization of 250 ?
     
  4. Oct 13, 2016 #3
    2, 5
     
  5. Oct 13, 2016 #4

    fresh_42

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    Consider the integer factorization. And how else can you write ##\frac{1}{250}##?
     
  6. Oct 13, 2016 #5
    Maybe this: [tex]5^{3} \cdot 2[/tex]

    I don't know how to flip it so it's 1/250. That's why I separated them.
     
  7. Oct 13, 2016 #6

    fresh_42

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    Better. And now ##\frac{1}{5^3\,\cdot\,2}## and your rules can be applied.
     
  8. Oct 13, 2016 #7

    SammyS

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    No. Those are the prime factors, but 2⋅5 = 10, not 250.

    Do a factor tree.

    How many factors of 2 and how many factors of 5 ?

    Added in Edit:

    I see you have it now.
     
  9. Oct 13, 2016 #8

    fresh_42

    Staff: Mentor

    Not that important. You can use rule two as well. Just know what ##\log_5 1## and ##\log_5 5## is.
     
  10. Oct 13, 2016 #9
    Okay thanks guys, so...

    [tex]\dfrac{1}{{5^{3}}\cdot2}[/tex]
    = [tex]log_{5}5^{3}-log_{5}5^{2}[/tex]
    = [tex]3log_{5}5 - 2log_{5}5[/tex]

    I'm closer, but still stuck on getting the correct answer, is this good so far?
     
  11. Oct 13, 2016 #10

    fresh_42

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    That's not what the quotient rule says, and what you did right the first time.
    Apply them one at a time: quotient rule - product rule - special cases. And don't forget parentheses when in doubt.
     
  12. Oct 14, 2016 #11
    Ok, so I see an error. I have a 5 base 5 in the second half of step 2. I think it should be [tex]log_{5}5^{3}-log_{5}2[/tex], then [tex]3log_{5}5^{}-log_{5}2[/tex]. Okay, but how do I get rid of the 5 and have a negative sign in front of the 3?
     
  13. Oct 14, 2016 #12

    SammyS

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    Your mistakes are in the quoted step.

    You used the quotient rule correctly in the OP, when you obtained
    ## \log_{5}1 - \log_{5}250 ##​
    Now use the factorization of 250: 2⋅53 to deconstruct log5(250).

    Notice:
    2 is not an exponent. It's a coefficient.​
    .
     
  14. Oct 14, 2016 #13
    Okay, so I have this so far: [tex]log_{5}1-log_{5}250[/tex] [tex]log_{5}1-log_{5}(5^{3}\cdot2)[/tex] [tex]log_{5}5^{3}-log_{5}2[/tex] [tex]3log_{5}5-log_{5}2[/tex]
     
  15. Oct 14, 2016 #14

    SammyS

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    Look at ##\ \log_5(250) = ... \ ## by itself. Then consider that it all comes from the denominator.
     
  16. Oct 14, 2016 #15
    Does that mean if it comes from denominator the whole solution becomes negative?
     
  17. Oct 14, 2016 #16

    Mark44

    Staff: Mentor

    The line above should be ##\log_5 \frac{1}{{5^{3}}\cdot2}##
    No. The thing you're taking the log of is a quotient that also includes a product.
    You're using [tex] tags, which causes the rendered expressions to be centered on their own line. For these equations, [itex] (inline LaTeX) would look better, IMO. It's simpler to surround the material with $$ (standalone) or ## (inline) tags.
     
  18. Oct 14, 2016 #17
    Okay, guys. This is as far as my knowledge takes me. [itex]log_{5}\dfrac{1}{250}[/itex]
    [itex]log_{5}1-log_{5}(5^{3}\cdot2)[/itex]
    [itex]log_{5}1-log_{5}5^{3}+log_{5}2[/itex]
    [itex]log_{5}1-3log_{5}5+log_{5}2[/itex] (log 5 base 5 cancels out)
    [itex]-3+log_{5}2[/itex]
    So which step did I go wrong? Also, thanks for the information, I was wondering how to format it.
     
  19. Oct 14, 2016 #18

    Mark44

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    Above is fine, but why not simplify it instead of dragging along ##\log_5(1)##? Do you know what this equals?
    The work below contains an error. The fact that you're dragging along ##\log_5(1)## probably contributed to this error.
     
  20. Oct 14, 2016 #19
    Doesn't [itex]log_{5}1[/itex] equal zero? I ignored it when solving the rest of the equation.
     
  21. Oct 14, 2016 #20

    Mark44

    Staff: Mentor

    Yes, that's the right value.

    Here's the work you showed in post#17.
    Dragging ##\log_5(1)## along in each of these steps probably caused the error I mentioned.

    BTW you are not "solving the equation." You are evaluating an expression; namely ##\log_5(\frac 1 {250})##. Solving an equation and evaluating or simplifying an expression are two completely different concepts.

    Here's a better start to this problem:
    ##\log_5(\frac 1 {250})##
    ##= \log_5(1) - \log_5(250)##
    ## = -\log_5(250)##
    ## = \dots ##
     
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