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How do you solve $$3^{x^2}=9^{4^x}+6\ ?$$ It seems really tough for me to find the value of $x$. What is the easy way to solve this kind of equations?

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  • $\begingroup$ "this kind of equations" is very hard to solve for real values of $x$ (even determining the number of solutions can be challenging). $\endgroup$ – user228113 Oct 12 '16 at 12:56
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    $\begingroup$ It hasn't got reasonable solutions...There's a solution near $-1.34986$, but that's just via numerical methods. $\endgroup$ – lulu Oct 12 '16 at 12:56
  • $\begingroup$ If you have a quora account, you could just have a look: quora.com/How-do-you-solve-3-x-2-9-4x-+6. It is a popular question. $\endgroup$ – StubbornAtom Oct 12 '16 at 12:58
  • $\begingroup$ @StubbornAtom That's $9^{4x}$...very different. Though it's a weird coincidence, perhaps the OP meant your version. $\endgroup$ – lulu Oct 12 '16 at 13:02
  • $\begingroup$ @lulu Yes I overlooked that. $\endgroup$ – StubbornAtom Oct 12 '16 at 13:05
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A good starting point is the factorization of non prime bases of powers:

$$3^{x^2}=(3^2)^{(2^2)^x}+6$$

This is

$$3^{x^2}=3^{2^{2x+1}}+6$$

Now, this equation can not be solved analitically. But we can use number theory to look for integer or rational solutions.

If $x$ is a positive integer, $3^{2^{2x+1}}$ is certainly a multiple of $9$, and $6$ is not. Therefore, $3^{x^2}$ is not a multiple of $9$. This means that $x$ must be $1$. But $x=1$ does not satisfy the equation.

It is easy to see that $x=0$ is not a solution.

If $x$ is a negative integer, LHS is integer, but RHS is irrational. So there is no integer solution.

If $x$ is not integer but $2x$ is integer, then the LHS is irrational but the RHS is rational.

More advanced number theory allows to claim that if $x$ is rational and $2x$ is not integer, $3^{x^2}$ is algebraic and $3^{2^{2x+1}}$ is trascendental. Then, the equation has no rational solution.

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I tried a similar line of attack that ajotatxe suggested, but could not get all the bases of 3 on one side so that a logarithm could simiplify things down.

I think this equation, as far as getting a practical solution, must be solved using some kind of root-finding method, e.g. Newton's Method, regula falsi, or any number of other methods.

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