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Math Help - Umm...exponent/square root problem

  1. #1
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    Umm...exponent/square root problem

    Okay Im trying to find log 5^x =5*sq rt of 5. Thats log base 5 with exponent x equals 5 times the square root of five or, with a base of five, what exponent do I need to get 5 times the square root of 5? Now I know to solve for an exponent you take the natural logarithm of both sides and solve for x which is easy and I know that the square root is the same thing as a number raised to the 1/2 power but even though I know how to solve components of this problem I dont know how to solve the actual problem.

    Okay let me try this: \log_{5}x=5\sqrt5
    Yes thats it, and I need to solve for x.
    Last edited by mgrade2000; February 21st 2010 at 04:03 PM.
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  2. #2
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    Quote Originally Posted by mgrade2000 View Post
    Okay Im trying to find log 5^x =5*sq rt of 5. Thats log base 5 with exponent x equals 5 times the square root of five or, with a base of five, what exponent do I need to get 5 times the square root of 5? Now I know to solve for an exponent you take the natural logarithm of both sides and solve for x which is easy and I know that the square root is the same thing as a number raised to the 1/2 power but even though I know how to solve components of this problem I dont know how to solve the actual problem.

    Okay let me try this: \log_{5}x=5\sqrt5
    Yes thats it, and I need to solve for x.
    the way you've written it ...

    \log_{5}x=5\sqrt{5}

    ... translates to

    5^{5\sqrt{5}} = x


    are you sure you don't mean

    \log_5(5\sqrt{5}) = x

    where ...

    5^x = 5\sqrt{5} ???
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  3. #3
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    No, I think I wrote it the right way and I think I figured it out:
    log_{5}x=5\sqrt{5}
    \sqrt{5} changes to 5^{1/2}
    Now I have log_{5}x=5*5^{1/2}
    This is asking what exponent do I raise a base of 5 to in order to get 5\sqrt{5} or 5*5^{1/2}
    to solve this:
    5^x= 5*5^{1/2}
    ln5^x= ln(5*5^{1/2})
    x(ln5)= ln(5*5^{1/2})
    x= [ln(5*5^{1/2})]/ln5
    x=ln11.18/ln5
    x=1.5
    Last edited by mgrade2000; February 21st 2010 at 05:05 PM.
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  4. #4
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    Quote Originally Posted by mgrade2000 View Post
    No, I think I wrote it the right way and I think I figured it out:
    log_{5}x=5\sqrt{5}
    \sqrt{5} changes to 5^{1/2}
    Now I have log_{5}x=5*5^{1/2}
    This is asking what exponent do I raise a base of 5 to in order to get 5\sqrt{5} or 5*5^{1/2}
    to solve this:
    5^x= 5*5^{1/2}
    ln5^x= ln(5*5^{1/2})
    x(ln5)= ln(5*5^{1/2})
    x= [ln(5*5^{1/2})]/ln5
    x=ln11.18/ln5
    x=1.5
    you did not write the log equation correctly.

    you worked it as ...

    \log_5(5\sqrt{5}) = x

    change to an exponential equation ...

    5^x = 5\sqrt{5} = 5^1 \cdot 5^{\frac{1}{2}} = 5^{\frac{3}{2}}

    x = \frac{3}{2}
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  5. #5
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    thanks a lot
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