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Is there a way to solve the following equation for $x$ in exact form $$(\frac{1}{2})^x + (\frac{1}{3})^x + (\frac{1}{4})^x = 1?$$

I've tried making the equation in this form $$(\frac{1}{2})^x + (\frac{1}{2})^{\log(3)x} + (\frac{1}{2})^{2x} = 1$$

and substituting $y = (1/2) ^ x$ to get

$$y^2 + y^{\log(3)} + y = 1,$$

which didn't help progress the calculation. I've done it numerically and it came out $1.0821...$ but I was wondering if there is an exact form to the solution.

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No, there's no exact solution to this equation because of the $(\frac{1}{3})^x$ term.

Without that there, it's easy, because you can transform it by a change of variables into a quadratic in terms of $2^{-x}$. But $3$ is not an integer power of $2$ and this makes this procedure impossible.

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    $\begingroup$ it is approximate $$1.0821314981404$$ with a numerical method e.g. the Newton method $\endgroup$ – Dr. Sonnhard Graubner Mar 5 '16 at 11:41
  • $\begingroup$ @Dr.SonnhardGraubner Yes the question asker has already got that numerically. $\endgroup$ – Deepak Mar 5 '16 at 11:42
  • $\begingroup$ What about in terms of log? I tried taking log of both sides but ended up with a log of a summation on the LHS. Anyway to solve this? $\endgroup$ – Syafiq Kamarul Azman Mar 5 '16 at 11:48
  • $\begingroup$ Sorry but many similar problems DO not have analitical solution, that is, you can not get the expression with finite combination of fundamental functions. $\endgroup$ – J. Yu Mar 5 '16 at 11:51
  • $\begingroup$ @J.Yu Can you provide another example? Perhaps along the same line as the question or with a proof for this? Thanks $\endgroup$ – Syafiq Kamarul Azman Mar 5 '16 at 12:32
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This is not an answer but it is too long for a comment.

As Deepak answered, there is no closed form because of the middle term and numerical solution must be considered (Newton being probably the simplest to use).

There is a interesting point here. Suppose that, probably as you did, we apply Newton method to $$f(x)=(\frac{1}{2})^x + (\frac{1}{3})^x + (\frac{1}{4})^x - 1$$ starting using $x_0=0$; the successive iterates will be $$x_1=0.629316$$ $$x_2=0.987611$$ $$x_3=1.077560$$ $$x_4=1.082120$$ $$x_5=1.082131$$ which is the solution for fifteen significant decimal places.

Now, let us do the same to $$g(x)=\log\Big((\frac{1}{2})^x + (\frac{1}{3})^x + (\frac{1}{4})^x\Big)$$ The successive iterates will be $$x_1=1.037061$$ $$x_2=1.082048$$ $$x_3=1.082131$$

Very different paths to the solution, isn't it ?

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