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I'm trying to use Wolfram|Alpha to solve a simple system of 4 non-linear equations. However, when I type in the equations it only pops out an equation for one of my unknown variables in terms of the other three. This is what I am typing into wolfram:

$\begin{cases}0.5D-2^{0.44A}+2^{0.567B}=0\\0.5D+1.005C+2^{0.389A}-2^{0.643B}=0\\0.5C+0.194A+0.321B=0\\0.5C-0.221A-0.283B=0\end{cases}$

and I am only getting an output of $D=-2(1.005C+e^{0.269634A}-e^{0.445694B})$

Please let me know if there is anyway to get solutions for all 4 variables, thanks!!

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  • $\begingroup$ Whoops sorry, there is a variable that is varied in the second equation. The equations I posted here is the case where that variable was zero. The 2nd equation should read: t=0.5D+1.005C+2^(0.389A)-2^(0.643B) where t will be varied from 0-50 $\endgroup$ – sabahdo Apr 10 '17 at 1:44
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Making all the numbers rational to avoid losses of accuracy, the equations write $$-2^{11 a/25}+2^{567 b/1000}+\frac{d}{2}=0\tag1$$ $$2^{389 a/100}-2^{643 b/1000}+\frac{201 c}{200}+\frac{d}{2}-t=0\tag2$$ $$\frac{97 a}{500}+\frac{321 b}{1000}+\frac{c}{2}=0\tag 3$$ $$-\frac{221 a}{1000}-\frac{283 b}{1000}+\frac{c}{2}=0\tag 4$$ We can use $(2)$, $(3)$ and $(4)$ to eliminate $b,c,d$ expressing them as functions of $a$. This leads to $$b=-\frac{415 }{604}a$$ $$c=\frac{16039 }{302000}a$$ $$d=-\frac{3223839 }{30200000}a+2^{1-\frac{53369 a}{120800}}-2^{\frac{389 a}{100}+1}+2 t$$ Replacing the above in $(1)$ leads to $$t=\frac{3223839 }{60400000}a-2^{-53369 a/120800}-2^{-47061 a/120800}+2^{11 a/25}+2^{389 a/100}$$ that you need to solve for $a$ for a given value of $t$.

I think that the easiest is to compute which would be $t$ for a given $a$; then the following table $$\left( \begin{array}{cc} a & t \\ 0.0 & 0.0000 \\ 0.1 & 0.4026 \\ 0.2 & 0.9003 \\ 0.3 & 1.5229 \\ 0.4 & 2.3091 \\ 0.5 & 3.3101 \\ 0.6 & 4.5923 \\ 0.7 & 6.2430 \\ 0.8 & 8.3764 \\ 0.9 & 11.1421 \\ 1.0 & 14.7358 \\ 1.1 & 19.4140 \\ 1.2 & 25.5122 \\ 1.3 & 33.4700 \\ 1.4 & 43.8631 \\ 1.5 & 57.4449 \end{array} \right)$$

So, for a given value of $t$, interpolate in the table to get a rather good estimate of $a$ and apply Newton method to polish the root.

Let me try for $t=10$; looking at the table $a_0=0.85$ seems to be reasonable. Starting with it, Newton iterates would be $$\left( \begin{array}{cc} n & a_n \\ 0 & 0.85000000 \\ 1 & 0.86199565 \\ 2 & 0.86181182 \\ 3 & 0.86181177 \end{array} \right)$$

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By subtracting the last two equations,

$$0.415A+0.604B=0$$ or

$$B=-0.6871A.$$

Then $$1.005 C=0.8350A.$$

Next, the two first equation can be written as

$$aA+b^A+c^A=d^A+e^A$$ for some real positive constants.

Such an equation must be solved numerically, after root isolation. $A=0$ is an obvious solution but there might be others. A plot should tell you.

The other unknowns easily follow (in particular, $B=C=D=0$).

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  • $\begingroup$ Very similar to Claude's answer. $\endgroup$ – Yves Daoust Apr 10 '17 at 12:50

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