I don't know how to get x in this equation: $$2^x+3^x=7^x$$ I need solution with steps, if possible.
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- $\begingroup$ I doubt this has a sensible closed form solution. I see $x\approx 0.668849674\cdots$ by numerical methods. $\endgroup$lulu– lulu2019-09-13 23:27:22 +00:00Commented Sep 13, 2019 at 23:27
- $\begingroup$ Its good to say if you are looking for an exact solution, approximate solution or what? Which methods do you know? This helps driving answers you could use. $\endgroup$NoChance– NoChance2019-09-14 00:07:18 +00:00Commented Sep 14, 2019 at 0:07
- $\begingroup$ I guess he wants to isolate x $\endgroup$Donlans Donlans– Donlans Donlans2019-09-14 00:12:32 +00:00Commented Sep 14, 2019 at 0:12
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2 Answers
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1 Hint: Consider $f(x)=\left(\dfrac27\right)^x+\left(\dfrac37\right)^x$. Show that $f(x)$ is a decreasing function, and by itv theorem deduce that the equation $f(x)=1$ has a unique solution. Use numerical method to approach the solution.
- $\begingroup$ I am not sure there is a function in the OP. The OP provided just an equation. $\endgroup$NoChance– NoChance2019-09-14 00:05:23 +00:00Commented Sep 14, 2019 at 0:05
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As already said, a closed form solution more than likely does not exits and some numerical method will be required. Consider that you look for the zero of function $$f(x)=7^x-(2^x+3^x)$$ DINEDINE showed that there is a unique solution in $(0,1)$.
$f(x)$ varies very fast. So, consider instead $$g(x)=x\log(7)-\log(2^x+3^x)$$ If you plot it, it looks very much like a straight line and this is very good for Newton method for example.
To have the starting guess, using Taylor at $x=0$ would give $$g(x)=-\log (2)+x \left(\log(7)-\frac{\log (2)}{2}-\frac{\log (3)}{2}\right)+O\left(x^2\right)$$ Ignoring the higher order terms, then $$x_0=\frac{2 \log (2)}{2 \log (7)-\log (2)-\log (3)}$$ Now, Newton iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.660121050806671 \\ 1 & 0.668848170790283 \\ 2 & 0.668849673876579 \\ 3 & 0.668849673876623 \end{array} \right)$$
Computed for $100$ exact significant figures, inverse symbolic calculators do not identify this number. However, you could be amazed that it is very close ($20$ correct figures) to $$\frac{127 e^2-441 e-1279}{26 e^2-438 e-1303}$$
