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Math Help - Equation

  1. #1
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    Red face Equation

    Find the solutions of x and n in this equation.( x^2+615=2^n)
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  2. #2
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    Re: Equation

    Solving for x...

    \displaystyle \begin{align*} x^2 + 615 &= 2^n \\ x^2 &= 2^n - 615 \\ x &= \pm \sqrt{2^n - 615} \end{align*}

    Solving for n...

    \displaystyle \begin{align*} 2^n &= x^2 + 615 \\ n &= \log_2{\left(x^2 + 615\right)} \end{align*}
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  3. #3
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    Re: Equation

    No, I mean the definite solution. I can get one solution from graph,but there exists no other solution. How can I prove it?
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  4. #4
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    Re: Equation

    Hello, Swarnav!

    \text{Find the solutions of }x\text{ and }n\text{ in this equation: }\:x^2+615\:=\:2^n

    Are there any restrictions on x and n?


    \text{Solving for }x\!:\;\;x \;=\;\pm\sqrt{2^n-615}

    \text{If }x\text{ must be real: }\;2^n - 615 \:\ge\;0 \quad\Rightarrow\quad n \:\ge\:\tfrac{\ln615}{\ln2} \:=\:9.2644426...


    I don't see any difficulty . . .

    \text{If }x = 9.3\text{, then }x = \sqrt{15.34593963...}

    \text{If }n = 10\text{, then }x = \sqrt{409}

    \text{If }n = 11\text{, then }x = \sqrt{1433}

    . . \text{ . . . and so on . . .}


    \text{If you insist that }x\text{ and }n\text{ are both } integers,
    . . \text{then we have a }big\text{ problem.}


    You say you found one solution graphically.
    I suspect that you graphed: . y \:=\: x^2+615\,\text{ and }\,y \,=\, 2^x
    . . and found their intersection.
    In doing so, you assumed that x = n.
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  5. #5
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    Re: Equation

    Yes x and n are both integers. Sorry I forgot to mention it. For this reason I cannot find solution of x. But there is one solution,if n=12 then we can get x as a integer.
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