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Logarithms where you are solving for X

  1. May 11, 2009 #1

    chemistudent

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

    Put in Exponential form and solve for X. SHOW YOUR STEPS!

    log2(x-5)=5

    2. Relevant equations

    I am not sure what this is asking for. There are no other equations.

    3. The attempt at a solution

    25=X-5

    25+5=X-5+5

    32+5=X

    X=37
     
    chemistudent, May 11, 2009
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  3. May 11, 2009 #2

    Dick

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    Looks good to me. What's the question?
     
    Dick, May 11, 2009
  4. May 11, 2009 #3

    chemistudent

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    How can I check this with a calculator? Because mine only has log10

    And if there isn't a way to check this with a calculator, Is there some website or something that i can use? Because when I look I can't find one.
     
    Last edited: May 11, 2009
    chemistudent, May 11, 2009
  5. May 11, 2009 #4

    Dick

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    You should be able to take log2(32) in your head. Since 32=2^5. But for future reference, log2(x)=log10(x)/log10(2).
     
    Dick, May 11, 2009
  6. May 11, 2009 #5

    chemistudent

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    um. I entered it into my calculator and I got an answer of 166.7025077 instead of 5.

    I entered (log(37)/log(2))(X-5)
     
    chemistudent, May 11, 2009
  7. May 11, 2009 #6

    Dick

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    Put x=37. log2(x-5)=log2(37-5)=log2(32)=log10(32)/log10(2)=1.50515/0.30103=5. But again, this isn't a question where you need to rely on a calculator.
     
    Dick, May 11, 2009
  8. May 11, 2009 #7

    chemistudent

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    ok then.

    um... one more question

    3(X+6)=27

    log327=X+6

    Aren't these the same thing?

    And if they are is this how you would answer this question?

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

    Put in Exponential form and solve for X. SHOW YOUR STEPS!

    3(X+6)=27

    log327=X+6

    2. Relevant equations

    I am not sure what this is asking for. There are no other equations.

    3. The attempt at a solution

    (log(27)/log(3))-6=X+6-6

    3-6=X

    X=-3
     
    chemistudent, May 11, 2009
  9. May 12, 2009 #8

    Mark44

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    Yes, the two equations are equivalent, which means essentially that they are both saying the same thing. Both have the same value of x as their solution.

    In the first equation, 3x + 6 = 27, x + 6 represents the exponent on 3 that gives you 27. That means that x + 6 = 3, or equivalently, that x = -3.


    In the second equation, log3 27 = x + 6, the left side simplifies to 3, giving you 3 = x + 6, or equivalently, x = -3.

    Every time you have an exponential equation, there is an equivalent log equation that has the same solution (as long as the base is positive and not equal to 1). The opposite is also true; namely that every log equation has an exponential equation counterpart.
     
    Mark44, May 12, 2009
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