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Math Help - How do you solve this logarithm question.

  1. #1
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    Question How do you solve this logarithm question.

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


    x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

    I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
    Last edited by mr fantastic; May 31st 2010 at 01:22 AM.
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    Quote Originally Posted by arijit2005 View Post
    FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -


    x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

    I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
    Dear arijit2005,

    Although you may not be familiar there's a logarithm identity,

    \huge{x^{log_{a}y}=y^{log_{a}x}}

    This will help you to solve the problem.
    Last edited by Sudharaka; May 31st 2010 at 04:38 PM.
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    Completely wrong here, see above and below posts.

    It might be prudent for this post to be deleted since it contributes nothing
    Last edited by e^(i*pi); May 31st 2010 at 01:24 AM.
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    Quote Originally Posted by e^(i*pi) View Post
    {\color{red}\log_2(a) \cdot \ln(x) + \log_2(x) \cdot \ln(a) = \ln(2) + 2\ln(a)}

    By the change of base rule: \log_2(x) = \frac{\ln(x)}{\ln(2)}

    \frac{\ln(a)}{\ln(2)} \cdot \ln(x) + \frac{\ln(x)}{\ln(2)} \cdot \ln(a) = \ln(2)+2\ln(a)

    \ln(a)\ln(x) + \ln(x)\ln(a) = \ln(2)(\ln(2)+2\ln(a))

    2\ln(a)\ln(x) = (\ln(2))^2 + 2\ln(a)\ln(2)

    \ln(x) = \frac{(\ln(2))^2 + 2\ln(a)\ln(2)}{2\ln(a)}


    Should be easy enough to find x in terms of a from there. I have used base e but any base will work
    Dear e^(i*pi),

    There's a mistake in your first step; log(A+B)\neq{logA+logB}
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    Quote Originally Posted by arijit2005 View Post
    FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -


    x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

    I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
    Quote Originally Posted by Sudharaka View Post
    Dear e^(i*pi),

    There's a mistake in your first step; log(A+B)\neq{logA+logB}
    Of course you're right, how on Earth did I not spot that >.<
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    Hello arijit2005
    Quote Originally Posted by arijit2005 View Post
    FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -


    x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

    I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
    We have to solve for x:
    x^{\log_2a}+a^{\log_2x}=2a^2
    If in doubt, get rid of logs as soon as you can - they're nasty things! So let b = \log_2a. Then a=2^b. So the equation becomes:
    x^b + \big(2^b\big)^{\log_2x} = 2\big(2^b\big)^2

    \Rightarrow x^b + 2^{b\log_2x} = 2\cdot2^{2b}

    \Rightarrow x^b + 2^{\log_2(x^b)} = 2\cdot2^{2b}

    \Rightarrow x^b + x^b = 2\cdot(2^2)^b

    \Rightarrow 2x^b = 2\cdot4^b

    \Rightarrow x = 4
    Grandad
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    Hi everyone,

    Another approch:

    If you use the logarithmic identity that I had mentioned in my previous post; \huge{x^{log_{a}y}=y^{log_{a}x}}

    x^{\log_2a}+a^{\log_2x}=2a^2

    \Rightarrow{a^{\log_2x}+a^{\log_2x}=2a^2}

    \Rightarrow{2a^{\log_2x}=2a^2}

    \Rightarrow{a^{\log_2x}=a^2}

    \Rightarrow{\log_2x=2}

    \Rightarrow{x=4}

    Hope this will help you.
    Last edited by Sudharaka; May 31st 2010 at 04:37 PM.
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    Quote Originally Posted by Grandad View Post
    Hello arijit2005We have to solve for x:
    x^{\log_2a}+a^{\log_2x}=2a^2
    If in doubt, get rid of logs as soon as you can - they're nasty things! So let b = \log_2a. Then a=2^b. So the equation becomes:
    x^b + \big(2^b\big)^{\log_2x} = 2\big(2^b\big)^2

    \Rightarrow x^b + 2^{b\log_2x} = 2\cdot2^{2b}

    \Rightarrow x^b + 2^{\log_2(x^b)} = 2\cdot2^{2b}

    \Rightarrow x^b + x^b = 2\cdot(2^2)^b

    \Rightarrow 2x^b = 2\cdot4^b

    \Rightarrow x = 4
    Grandad

    WOW.. Thanks.. That's a nice way.. Why the hell didn't I think of that??? I need more practice.
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  9. #9
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    Quote Originally Posted by Sudharaka View Post
    Hi everyone,

    Another approch:

    If you use the logarithmic identity that I had mentioned in my previous post; \huge{x^{log_{a}y}=y^{log_{b}x}}

    x^{\log_2a}+a^{\log_2x}=2a^2

    \Rightarrow{a^{\log_2x}+a^{\log_2x}=2a^2}

    \Rightarrow{2a^{\log_2x}=2a^2}

    \Rightarrow{a^{\log_2x}=a^2}

    \Rightarrow{\log_2x=2}

    \Rightarrow{x=4}

    Hope this will help you.

    Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula \huge{x^{log_{a}y}=y^{log_{b}x}}

    By the way, is it \huge{x^{log_{a}y}=y^{log_{b}x}} or \huge{x^{log_{a}y}=y^{log_{a}x}}???

    Will base "a" change?? You actually wrote \huge{y^{log_{b}x}}

    Are you sure the a becomes b???
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  10. #10
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    Hello arijit2005
    Quote Originally Posted by arijit2005 View Post
    Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula \huge{x^{log_{a}y}=y^{log_{b}x}}

    By the way, is it \huge{x^{log_{a}y}=y^{log_{b}x}} or \huge{x^{log_{a}y}=y^{log_{a}x}}???

    Will base "a" change?? You actually wrote \huge{y^{log_{b}x}}

    Are you sure the a becomes b???
    Yes, it should be
    x^{\log_ay}=y^{\log_ax}
    You can prove this using the same substitution that I used (so that you can get rid of those nasty logs!):
    Let b = \log_ay and c = \log_ax

    Then a^b = y and a^c = x

    So x^{\log_ay} = x^b
    =(a^c)^b

    = (a^b)^c

    = y^c

    =y^{\log_ax}
    Grandad
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