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Math Help - x=(1.5)^x

  1. #1
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    x=(1.5)^x

    Hi,
    I don't remember how can I solve it for example: x=1,5^x
    Help me please,
    The Riddler
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  2. #2
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    I assume this equation is \displaystyle x = 1.5^x. You can't solve it exactly, you have to use a numerical method.
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  3. #3
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    Quote Originally Posted by Riddler View Post
    Hi,
    I don't remember how can I solve it for example: x=1,5^x
    Help me please,
    The Riddler
    This equation doesn't have a real solution because (1.5)^x > x\ ,\ x \in \mathbb{R}

    I've checked at which x-value the function f(x)=(1.5)^x has the slope 1. I got the point

    P\left(\dfrac{\ln\left( \frac1{\ln(1.5)} \right)}{\ln(1.5)}~,~\dfrac1{\ln(1.5)} \right)

    and this point lies above the line y = x.
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  4. #4
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    I will admit that I would have interpreted the question as "if x= 1, what is 5^x", but solving the equation x= 1.5^x is much more interesting.

    Of course, 1.5^x= e^{ln(1.5^x)= e^{xln(1.5)} so x= 1.5^x is the same as x= e^{xln(1.5)} and that is the same as xe^{x(-ln(1.5))}= 1. Let y -ln(1.5)x so that x= \frac{y}{-ln(1.5)} and the equation becomes \frac{y}{-ln(1.5)}e^y= 1 or ye^y}= -ln(1.5).

    Now, we have y= W(-ln(1.5)) where W is "Lambert's W function", the inverse function to [itex]f(x)= xe^x[/itex].

    Finally, then x= \frac{y}{-ln(1.5)}= \frac{W(-ln(1.5))}{-ln(1.5)}.

    Of course, as earboth says, x is not a real number because the domain of the real valued W function, while it includes some negative numbers, does not include any that negative.
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  5. #5
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    My previous example was wrong.
    But if: x=(1.15)^x
    It has got a real solution: x=22.1723..
    How can I solve it?
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  6. #6
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    It also has a solution of x \approx 1.18

    Not sure how you'd show it though, I used wolfram
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  7. #7
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    Drawing the graphs of \displaystyle y = x and \displaystyle y = 1.15^x and noting where they intersect gives you an idea of the solutions...
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  8. #8
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    Of course, you could use any number of methods of approximation. I tried the risky approach of blind iteration, using the following steps (having drawn the graphs first for a rough idea of the roots, obviously):
    x_{n+1}=1.15^{x_n}


    Starting at x_1=20

    x_2=1.15^{20}=16.36

    x_3=1.15^{16.36}=9.85

    x_4=1.15^{9.85}=3.96
    ....
    We get the x\approx~1.18 solution at approximately the 9th iteration. However I was trying to get the other solution! You could try a slightly better approach, such as interval bisection, Newton-Raphson or linear interpolation or something, but then you probably know all this already.
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  9. #9
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    I wrote a program yesterday, which has solved my problem.
    You can download it here: http://strategist.fw.hu/program.rar
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