Results 1 to 10 of 10

Math Help - How do you solve this logarithm question.

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    18

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    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}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    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 >.<
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    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
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2009
    From
    1111
    Posts
    872
    Thanks
    3
    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.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2010
    Posts
    18
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    May 2010
    Posts
    18
    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???
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor
    Grandad's Avatar
    Joined
    Dec 2008
    From
    South Coast of England
    Posts
    2,570
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How do you solve this logarithm?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 14th 2010, 07:14 AM
  2. Solve the logarithm.
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 22nd 2010, 01:12 PM
  3. Logarithm Help: Solve for x.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 1st 2009, 11:12 PM
  4. Solve for x logarithm
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 17th 2008, 07:10 AM
  5. Solve Logarithm
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 28th 2008, 12:44 PM

Search Tags


/mathhelpforum @mathhelpforum