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Changing the base logarithms homework

  1. Oct 3, 2005 #1
    log_2(x) + log_4(x) = 2

    I've tried everything i can think of, including changing the base - to no avail.

    Any ideas?!
     
  2. jcsd
  3. Oct 3, 2005 #2

    TD

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    It's doable by changing base, but you can also do:

    [tex]\log _2 x + \log _4 x = 2 \Leftrightarrow 4^{\log _2 x + \log _4 x} = 4^2 \Leftrightarrow 4^{\log _2 x} 4^{\log _4 x} = 16 \Leftrightarrow x^2 \cdot x = 16[/tex]
     
  4. Oct 3, 2005 #3
    Can you explain those steps to me because I don't really understand
     
  5. Oct 3, 2005 #4

    TD

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    - We raise both sides as exponents of base 4.
    - Then we use the fact that [itex]a^{b + c} = a^b \cdot a^c[/itex]
    - Then we can simplify the powers and logs because they are inverse operation.

    Please specify what steps you have trouble with.

    You can also do:

    [tex]\log _2 x + \log _4 x = 2 \Leftrightarrow \log _4 x^2 + \log _4 x = 2 \Leftrightarrow \log _4 \left( {x \cdot x^2 } \right) = 2 \Leftrightarrow \log _4 \left( {x^3 } \right) = 2[/tex]

    And then take 4^ again of both sides.
     
  6. Oct 3, 2005 #5

    arildno

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    Here's how you can change base:
    By definition, for y>0
    [tex]y=2^{\log_{2}y}=4^{\log_{4}y}[/tex]
    Thus, from the last identity, we get:
    [tex]\log_{2}y\log_{4}2=\log_{4}y[/tex]
    But:
    [tex]2=4^{\frac{1}{2}}[/tex]
    thus:
    [tex]\log_{4}2=\log_{4}4^{\frac{1}{2}}=\frac{1}{2}[/tex]
    or:
    [tex]\log_{2}y=2\log_{4}y[/tex]
    Thus, setting y=x, we have:
    [tex]\log_{2}x+\log_{4}x=3\log_{4}x=\log_{4}x^{3}[/tex]
    Your original equation is therefore:
    [tex]\log_{4}x^{3}=2[/tex]
    Or:
    [tex]x^{3}=4^{2}[/tex]
    which agrees with Tide's answer..
     
  7. Oct 3, 2005 #6

    TD

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    Hmm, I've seen a Tide here, but it's not me :biggrin:
     
  8. Oct 3, 2005 #7
    This is going to sound pathetic so i apologise but i do not know how these steps work!

    [tex]\log _2 x + \log _4 x = 2 \Leftrightarrow 4^{\log _2 x + \log _4 x} = 4^2[/tex]

    [tex]4^{\log _2 x} 4^{\log _4 x} = 16 \Leftrightarrow x^2 \cdot x = 16[/tex]

    I'm really sorry but can you give me an anser for first principles - this isn't a requirement of the questoin but i actually want to understand this!
     
  9. Oct 3, 2005 #8

    arildno

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    Allright tiddely-doo, I made a mistake okay!! :cry:
    I won't do it again.. :blushing:
     
  10. Oct 3, 2005 #9

    TD

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    Well, we use that:
    [tex]\begin{gathered}
    a^{\log _a x} = x \hfill \\
    a^{b + c} = a^b \cdot a^c \hfill \\
    \end{gathered} [/tex]
     
  11. Oct 3, 2005 #10
    i didn't know the result a^log_a(x) = x
     
  12. Oct 3, 2005 #11

    TD

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    That is because they are by definition inverse operations.

    [tex]a^{\log _a x} = x \Leftrightarrow \log _a \left( {a^{\log _a x} } \right) = \log _a x \Leftrightarrow \log _a x \cdot \log _a a = \log _a x \Leftrightarrow \log _a x = \log _a x[/tex]

    So it is indeed correct.
     
  13. Oct 4, 2005 #12
    you probably know the formula
    [tex]log_a (x^n) = n log_a (x)[/tex], where [tex]x>0[/tex]

    There is a very useful similar formula:
    [tex]log_{a^n} (x) = (1/n) log_a (x)[/tex]
    (can be derived from [tex]log_a (x) = 1/ log_x (a)[/tex] ).

    For your equation
    [tex]log_2 (x) + log_4 (x) = 2[/tex]
    [tex]log_2 (x) + log_{2^2} (x) = 2[/tex]
    [tex]log_2 (x) + (1/2) log_2 (x) = 2[/tex]
    [tex](3/2) log_2 (x) = 2[/tex]
    [tex]log_2 (x) = 4/3[/tex]
    [tex]x = 2^{4/3}[/tex]
     
    Last edited: Oct 4, 2005
  14. Oct 4, 2005 #13
    TD, actually this property is just a definition of logarithm:
    [tex]{\log_a x}[/tex] is such a number [tex]n[/tex] that
    [tex]a^{n} = x[/tex]
    Substituting [tex]{\log_a x}[/tex] for [tex]n[/tex]
    [tex]a^{\log_a x} = x[/tex]
     
    Last edited: Oct 4, 2005
  15. Oct 4, 2005 #14

    TD

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    Which is why I said "they are by definition inverse operations".

    What came after that was just a way of using logarithm properties he might know to show that it is equal.
     
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