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If the log is expressed with base 9 it is equal to:

  1. Apr 23, 2015 #1
    1. The problem statement, all variables and given/known data
    If log3(x-3)-log1/3(x+1) is expressed in terms of a logarithm with base 9, then it is equal to?

    2. Relevant equations
    logba=logca/logcb
    3. The attempt at a solution
    I know I need to get rid of the 1/2 by multiplying but can't remember how to apply it.

    log3(x-3)-log1/3
    =log9(x-3) / log93 - log9(x+1) / log91 / 3
    = log9(x-3) / 1/2 - log9(x+1) / -1/2

    Can someone tell me how to display fractions correctly?
    Thank you.

    Jaco
     
    Last edited: Apr 23, 2015
  2. jcsd
  3. Apr 23, 2015 #2

    Suraj M

    User Avatar
    Gold Member

    where'd you get that ##(x + 1)## i dont see it in the question..
    Well, you finished the hard part,
    i might be wrong , but can't you just takr ##2## common and express the the 2 term expression in terms of ##log_9\frac{numerator}{denominator}##
    if you're referring to typing like this..
    $$\frac{log_9(x-3)}{\frac{1}{2}} - \frac{log_9(x+1)}{\frac{-1}{2}}$$this might help... go through this https://www.physicsforums.com/threads/physics-forums-faq-and-howto.617567/#post-3977517
     
    Last edited: Apr 23, 2015
  4. Apr 23, 2015 #3
    Its in the question, I struggle to read in the code, sorry about that

    This is what it feels like I should do, but I am not sure? would I add a 2 in from of the log?
    Yes thats what Im referring to, Thank you.
     
  5. Apr 23, 2015 #4

    Mark44

    Staff: Mentor

    The ⇒ means "implies", which isn't appropriate here. What you should have above is =. The expression on the first line is equal to the one on the second line, which in turn is equal to the one on the third line.

    The formula you show as your relevant equation I don't keep in memory. Instead I derive it when I need it, which takes a little longer, but so be it.

    Here's an example of how I would rewrite ##log_2(x + 5)## as a logarithm in base 4.
    Let ##y = log_2(x + 5)##
    Then ##x + 5 = 2^y## See comment 1 below.
    Taking the log (base 4) of both sides, I get
    ##log_4(x + 5) = log_4(2^y)##
    Or, ##log_4(x + 5) = y * log_4(2)##, using one of the properties of logs
    Solving for y, I get
    ##y = \frac{log_4(x + 5)}{log_4(2)} = \frac{log_4(x + 5)}{1/2} = 2 * log_4(x + 5)## See comment 2 below
    Then ##log_2(x + 5) = 2 * log_4(x + 5)##

    1. I used the most important concept in logarithms; that a logarithm is an exponent on whatever the base happens to be. For example, ##log_2(x + 5)## means the exponent on 2 (the base) the results in x + 5.
    2. In this step I replaced ##log_4(2)## with 1/2. ##log_4(2)## means the exponent on 4 that results in 2. In other words, ##4^? = 2##. The exponent has to be 1/2, since ##4^{1/2} = 2##.

     
  6. Apr 23, 2015 #5
    Is this not what I have done on mine?

    Could you explain the step above to me?*2 to remove the 1/2, but why does the 2 remain in front of the log?
    Because this is what I want to do but I am not sure why...[/QUOTE]
     
    Last edited: Apr 23, 2015
  7. Apr 23, 2015 #6

    Mark44

    Staff: Mentor

    I don't know what you mean. What are you asking here?
    Yes, this is what you did. You said that you needed to get rid of the 1/2 by multiplying, but couldn't remember how to do it. I wasn't sure if you knew what you were doing, or were just following a cookbook formula that you might not remember or might remember incorrectly. What I wrote was an explanation of what should happen.
    Divide by 1/2 is exactly the same as multiplying by the reciprocal of 1/2 -- in other words, the same as multiplying by 2.
     
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