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Thread: Solving this Logarithmic equation

  1. #1
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    Solving this Logarithmic equation

    I would like to know how much 'n' equals, based on the formula below:

    $\displaystyle 3^{log_2(n)} = X$
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  2. #2
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    Re: Solving this Logarithmic equation

    Quote Originally Posted by mohamedennahdi View Post
    I would like to know how much 'n' equals, based on the formula below:

    $\displaystyle 3^{log_2(n)} = X$
    Hint: Start by taking log base 3 of both sides.

    -Dan
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  3. #3
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    Re: Solving this Logarithmic equation

    What do you think of this?

    $\displaystyle \\3^{log_2(n)} = 3^{\frac{log_3(n)}{log_3(2)}}\\\\3^{log_2(n)} = n^{\frac{1}{log_3(2)}}\\\\3^{log_2(n)} = \sqrt[log_3(2)]{n}$
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  4. #4
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    Re: Solving this Logarithmic equation

    Quote Originally Posted by mohamedennahdi View Post
    What do you think of this?

    $\displaystyle \\3^{log_2(n)} = 3^{\frac{log_3(n)}{log_3(2)}}\\\\3^{log_2(n)} = n^{\frac{1}{log_3(2)}}\\\\3^{log_2(n)} = \sqrt[log_3(2)]{n}$
    You still haven't separated the n from one side.

    Here's where I was pointing you:
    $\displaystyle 3^{log_2(n)} = x$

    $\displaystyle log_3 \left ( 3^{log_2(n)} \right ) = log_3(x)$

    $\displaystyle log_2(n) \cdot log_3(3) = log_3(x)$

    Can you finish from here?

    -Dan
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  5. #5
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    Re: Solving this Logarithmic equation

    No, please proceed!
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  6. #6
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    Re: Solving this Logarithmic equation

    Quote Originally Posted by mohamedennahdi View Post
    No, please proceed!
    Aw c'mon! Are you telling me you can't solve an equation of the form a*log_2(n) = b? Give it a try! I'll even tell you that log_3(3) = 1.

    -Dan
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  7. #7
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    Re: Solving this Logarithmic equation

    How about this:

    $\displaystyle \\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$
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  8. #8
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    Re: Solving this Logarithmic equation

    Quote Originally Posted by mohamedennahdi View Post
    How about this:

    $\displaystyle \\log_2(n).log_3(3) = log_3(x)\\\\log_2(n) = log_3(x)\\\\n = log_3(x)^2\\\\n = 2log_3(x)$
    To get rid of a "log" you have to raise it to a power. So after line 2 we would have
    $\displaystyle log_2(n) = log_3(x)$

    $\displaystyle 2^{log_2(n)} = 2^{log_3(x)}$

    $\displaystyle n = 2^{log_3(x)}$

    -Dan
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