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Is there any way to solve the equation $3-x^2 = 2^x$ without using the calculator to estimate the solution for you? So far I've tried changing the base of the right hand side to base $e$ so $3-x^2 = e^{x\ln2}$ ,but the equation still cant be isolated for x?
$$\frac{\ln(3-x^2)}{x} = \ln2$$
Am I missing something crucial here?

Shraddheya Shendre
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Prandals
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  • $y=3-x^2$ and $y=2^x$ can be drawn using a graphing tool. By inspection one can see that $x=1$ is a solution. There is another solution too, but that one is "ugly".Without any devices, I am not aware that you can algebraically manipulate your equation to arrive at those two solutions. – imranfat Nov 28 '16 at 04:42
  • @imranfat: The required manipulation is very straightforward. One of the solutions will of course be expressed in terms of one or more logarithms of specific numbers, but that's fine. – Brian M. Scott Nov 28 '16 at 04:48
  • If that is allowed, sure then that makes sense. – imranfat Nov 28 '16 at 04:49
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    Do you mean $3-x^2$ or $3^{-x^2}$ on the left? imranfat's comment solves one version, Brian Scott's answer the other. – Ross Millikan Nov 28 '16 at 04:56
  • @ Ross Millikan it is 3 - [x^2] Sorry, but I do not know how to format it like you did – Prandals Nov 28 '16 at 04:58

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HINT: $\ln 3^{-x^2}=-x^2\ln 3$, and $\ln 2^x=x\ln 2$. Let $a=\ln 3$ and $b=\ln 2$; then after taking the natural log on both sides you have $-ax^2=bx$. Can you solve that for $x$?

Once you've found its solutions in terms of the constants $a$ and $b$, you can always substitute the correct numbers for $a$ and $b$.

Brian M. Scott
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