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Math Help - [SOLVED] Log Function

  1. #1
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    Red face [SOLVED] Log Function

    Real variables:

    ln(2+y) = 5ln(3-x) -2sqrtx

    (where sqrtx = the square root of x sorry!)

    I need to find the range of values of x and y for which they're defined and express as y = f(x)

    Can someone show me how to do these types of questions so I can do the rest? I'm stuck!!

    Many thanks, I appreciate it
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  2. #2
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    First of all to express as  y = f(x) take the exponential of both sides and then subtract 2 from both sides:

     \ln(2+y) = 5\ln(3-x) - 2\sqrt{x}
     2+y = e^{5\ln(3-x) - 2\sqrt{x}}
     y= e^{5\ln(3-x) - 2\sqrt{x}} - 2

    Then for the values for which x and y are defined you have to look at which values are not possible in the original equation:
     \ln(2+y) = 5\ln(3-x) - 2\sqrt{x}

    Dealing with values of x and first looking at the square root sign it must be true that  x\geq0 otherwise we would be taking the square root of a negative number which isn't allowed. Going on to look at  \ln(3-x) we must use values  x\leq 3 or else we would be taking the ln of a number >0 which does not exist.
    So the values of x for which the equation is defined would be:

     0 \leq x \leq 3 where x is a member of \Re

    Looking at values for y there is only one restriction shown by  \ln(2+y) in the same way to the earlier log it must be  \geq 0 so:

     y\geq -2 where y is a member of \Re

    If you have further questions just ask.
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  3. #3
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    Hello, MathLoser!

    This is a tricky one . . .


    \ln(2+y) \;= \;5\ln(3-x) -2\sqrt{x}

    Find the range of values of x and y for which they're defined
    and express as: y \,=\,f(x)

    From \sqrt{x}, we see that: x \geq 0

    From \ln(3-x), we see that: x < 3

    . . The domain is: . \boxed{\;0\: \leq\: x\: <\: 3\;}


    We have: . \ln(2+y) \:=\:\ln(3-x)^5 - 2\sqrt{x}

    Then: . y + 2 \;=\;e^{\ln(3-x)^5 - 2\sqrt{x}}  \;=\;e^{\ln(3-x)^5}\cdot e^{-2\sqrt{x}} \;=\;\frac{e^{\ln(3-x)^5}}{e^{2\sqrt{x}}} \;=\;\frac{(3-x)^5}{e^{2\sqrt{x}}}

    . . Hence: . \boxed{y \;=\;\frac{(3-x)^5}{e^{2\sqrt{x}}}-2}


    When x = 0\!:\;y \:=\:\frac{3^5}{e^0} - 2 \:=\:241

    When x = 1\!:\;\;y \:=\:\frac{2^5}{e^2} \:\approx\:2.331

    When x = 2\!:\;\;y \:=\:\frac{1^5}{2^{2\sqrt{2}}} \:\approx\:-1.982

    When x\to3\!:\;\;y \to -2


    The range seems to be: . \boxed{\;-2 \:<\: y \:\leq \:241\;}

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  4. #4
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    Red face

    Hi and thanks to both of you for your posts - do they agree or are there slightly different answers?
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