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Find all real solution of the equation $2^x+3^x+6^x = x^2$

$\bf{My\; Try::}$ Let $$f(x) = 2^x+3^x+6^x-x^2\;,$$ Now Using first Derivative

$$f'(x) = 2^x\cdot \ln 2+3^x\cdot \ln 3+6^x\cdot \ln 3-2x$$

Now for $x<0\;,$ We get $f'(x)>0,$ So function $f(x)$ is strictly Increasing function.

so it will cut $\bf{X-}$ axis at exactly one point.

So $f(x)=0$ has exactly one root for $x=-1$ for $x<0$

Now How can I calculate no. of real roots for $x\geq 0$

Thanks

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  • $\begingroup$ I think you may be able to make use of something like $2^x\geq x^2$ for $x>0$. This would imply there are no positive roots. I am not certain as I haven't worked it out, just a passing idea. $\endgroup$
    – Clayton
    Sep 13 '15 at 17:14
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    $\begingroup$ If it is increasing (actually in is increasing for $x \ge -1$) and you have found one root, you are right there meaning you've solved it, $\endgroup$
    – Shailesh
    Sep 13 '15 at 17:14
  • $\begingroup$ Shailesh is right, you've solved your problem. $\endgroup$
    – Victor
    Sep 13 '15 at 17:15
  • $\begingroup$ You have a typo in your f' $\endgroup$ Sep 14 '15 at 11:28
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The derivative $f'(x)$ is certainly greater than $h(x)=6^x \ln 6 - 2x.$ [I assume you had a typo when you wrote the term $6^x \ln 3,$ as the derivative of $6^x$ is $6^x \ln 6.$]

Now assume $x \ge 0$ and note $h(0)=\ln 6 >0.$ Also we have $$h'(x)=6^x (\ln 6)^2 -2 > (3.2)\cdot 6^x -2\ge 1.2 >0,$$ using the underestimate $3.2$ for $(\ln 6)^2$ and the fact that for nonnegative $x$ one has $6^x \ge 1.$

Thus we have both $h(0)>0$ and $h$ increasing on $[0,\infty),$ and can conclude as desired that $h(x) \ge 0$ for $x \ge 0.$ Combining this with your result for negative $x$ one gets $f$ increasing on $\mathbb{R}$ so the only zero of $f$ is at $x=-1.$

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I'm using the f(x) you defined. You can numerically verify that f(0) and f'(0) are both >0. You can analytically verify that all higher derivatives of f(x) will be >0 at x=0.

So if f(x) and all of it's derivatives are >0 at x=0, it is not possible for f(x) to dip down to 0 for any x>0

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  • $\begingroup$ This conclusion relies on $f$ being analytic with infinite radius of convergence, which is true here, but maybe should be mentioned. Otherwise consider $f(x)=e^x$ in a neighborhood of $0$ but patched in a $C^\infty$ way with another smooth function, to make it "dip down" below the $x$ axis for some positive $x$ $\endgroup$
    – coffeemath
    Sep 14 '15 at 14:50
  • $\begingroup$ @coffeemath None of that applies to this problem. $\endgroup$ Sep 14 '15 at 15:02
  • $\begingroup$ Jerry: But the original $f(x)$ of the problem is surely analytic with infinite radius of convergence. I'm just saying your "not possible" part of the argument doesn't work without assuming $f$ is analytic. $\endgroup$
    – coffeemath
    Sep 14 '15 at 15:06
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    $\begingroup$ @coffeemath I understand that, but I'm trying to not confuse the Question-asker with extra discussion and terms that might be unfamiliar. $\endgroup$ Sep 14 '15 at 15:24
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Your derivative is equal to

$f'(x) = \ln{ 2^{2^x}+ \ln 3^{3^x}+ \ln 6^{6^x}-\ln e^{-2x}=\ln2^{2^x}3^{3^x}6^{6^x}e^{2x}}$

Without calculating $f '(0)$, it is clear that it is positive and increasing so positive for $x\ge 0$. Hence $f$ is increasing for $x\gt 0$; besides $f(0)=3$ thus none positive solution.

NOTE.-A comment induced me to edit what was right for something wrong. Again I put what I wrote first.

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  • $\begingroup$ Shouldn't the last term be $-\ln e^{2x}=-2x$? (Still get $f'\ge 0$ for $x \ge 0$) $\endgroup$
    – coffeemath
    Sep 14 '15 at 3:14
  • $\begingroup$ Once one puts the $e^{2x}$ in the denominator (where I think it should go in this approach) it becomes less than immediate that the quantity one is taking the log of exceeds $1,$ which one needs to conclude $f'>0.$ It does exceed one, but that takes some nontrivial check. $\endgroup$
    – coffeemath
    Sep 14 '15 at 10:21
  • $\begingroup$ @coffeemath: Thank you very much. $\endgroup$
    – Piquito
    Sep 14 '15 at 12:54
  • $\begingroup$ It should still be changed, this time to $e^{2x}$ in the denominator rather than $e^{x^2}.$ Note that you need (a bit) more to conclude $f'>0,$ since you need what's inside the ln to exceed 1, not just exceed 0. $\endgroup$
    – coffeemath
    Sep 14 '15 at 14:35
  • $\begingroup$ If $f(x)=2^x+3^x+6^x-x^2$ as in the post, then $f'(x)=2^x \ln 2 +3^x \ln 3+6^x \ln 6 -2x$ (also as in post). Using log rules this is $\ln 2^{2^x}+\ln 3^{3^x}+\ln 6^{6^x}-2x.$ Since $2x=\ln e^{2x},$ in converting the subtracted $2x$ to a log term like the others, one ends up subtracting $\ln e^{2x},$ rather than subtracting $\ln e^{-2x}$ as you had it originally and have it now. Your first displayed version of $f'(x)$ ends up with $-\ln e^{-2x}=-(-2x)=+2x$ which is not the last term of the derivative of $f$, but has the wrong sign. $\endgroup$
    – coffeemath
    Sep 15 '15 at 13:35

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