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Thread: [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 $\displaystyle y = f(x) $ take the exponential of both sides and then subtract 2 from both sides:

    $\displaystyle \ln(2+y) = 5\ln(3-x) - 2\sqrt{x} $
    $\displaystyle 2+y = e^{5\ln(3-x) - 2\sqrt{x}} $
    $\displaystyle 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:
    $\displaystyle \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 $\displaystyle x\geq0 $ otherwise we would be taking the square root of a negative number which isn't allowed. Going on to look at $\displaystyle \ln(3-x) $ we must use values $\displaystyle x\leq 3 $ or else we would be taking the ln of a number $\displaystyle >0$ which does not exist.
    So the values of x for which the equation is defined would be:

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

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

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

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

    This is a tricky one . . .


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

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

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

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

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


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

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

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


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

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

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

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


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

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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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