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Math Help - Logarithm Problem?? How to find the value of "X"??

  1. #1
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    Exclamation Logarithm Problem?? How to find the value of "X"??

    Okay here's the problem: -

    FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

    [LOG2LOG2LOG2X = 1]

    The given answer is "X" = 16.

    But I have no idea how to find that out. Do I have to write LOG22 instead of 1?

    Please help...


    Thanks in advance...
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  2. #2
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    Quote Originally Posted by arijit2005 View Post
    Okay here's the problem: -

    FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

    [LOG2LOG2LOG2X = 1]

    The given answer is "X" = 16.

    But I have no idea how to find that out. Do I have to write LOG22 instead of 1?

    Please help...


    Thanks in advance...

    Work from the outside in.

    So log_2(log_2(log_2(x))) = 1

    log_2(y) = 1

    y = 2

    Now write

    log_2(log_2(x)) = 2

    log_2(z) = 2

    z = 4

    Lastly,

    log_2(x) = 4

    x = 16

    Alternatively, you could do LHS = RHS \Longrightarrow 2^{LHS} = 2^{RHS} three times. (LHS = left hand side, etc.)
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  3. #3
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    Quote Originally Posted by undefined View Post
    Work from the outside in.

    So log_2(log_2(log_2(x))) = 1

    log_2(y) = 1

    y = 2

    Now write

    log_2(log_2(x)) = 2

    log_2(z) = 2

    z = 4

    Lastly,

    log_2(x) = 4

    x = 16

    Alternatively, you could do LHS = RHS \Longrightarrow 2^{LHS} = 2^{RHS} three times. (LHS = left hand side, etc.)

    Thanks...

    I've thought of another way, but am not sure if it's valid. Please let me know if it's correct: -

    log_2(log_2(log_2(x))) = 1

    log_2(log_2(log_2(x))) = log_2(2) [I can write log_2(2) instead of 1, can't I?]

    So,

    log_2(log_2(x)) = 2 [Now this is what I'm unsure about. Can I eliminate log_2 from both sides in this case??]

    Next,

    log_2(log_2(x)) = 2*log_2(2) [where log_2(2) = 1 ]

    log_2(log_2(x)) = log_2(2)^2 \Longrightarrow log_2(log_2(x)) = log_2(4)

    log_2(x) = 4

    \Longrightarrow log_2(x) = 4*log_2(2)

    \Longrightarrow log_2(x) = log_2(2)^4

    \Longrightarrow log_2(x) = log_2(16)

    Thus x = 16

    Now is it a valid method???
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by arijit2005 View Post
    Thanks...

    I've thought of another way, but am not sure if it's valid. Please let me know if it's correct: -

    log_2(log_2(log_2(x))) = 1

    log_2(log_2(log_2(x))) = log_2(2) [I can write log_2(2) instead of 1, can't I?] Yes.

    So,

    log_2(log_2(x)) = 2 [Now this is what I'm unsure about. Can I eliminate log_2 from both sides in this case??]

    Yes you can. This is because the function \color{red}f(x)=\log_2(x) is one-to-one, or injective, which you probably haven't learned yet. But it means that f(a) = f(b) implies a = b, and graphically, f(x) passes the "horizontal line test."

    Next,

    log_2(log_2(x)) = 2*log_2(2) [where log_2(2) = 1 ]

    log_2(log_2(x)) = log_2(2)^2 \Longrightarrow log_2(log_2(x)) = log_2(4) I would write \color{red}log_2(2^2) so people don't misinterpret as \color{red}\left(\log_2(2)\right)^2.

    log_2(x) = 4

    \Longrightarrow log_2(x) = 4*log_2(2)

    \Longrightarrow log_2(x) = log_2(2)^4

    \Longrightarrow log_2(x) = log_2(16)

    Thus x = 16

    Now is it a valid method??? Yes. Good job.
    ..
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  5. #5
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    Thanks..

    No I haven't learned the " f(x) = log_2(x) " thing yet.
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