If $ x(9^{\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3}}) = (3^{2\sqrt{x^{2}-3}+1} - 3^{\sqrt{x^{2}-3} + 1} - 18) \sqrt{x}+ 6x $, what is the maximum $x$?

Question

If $$ x(9^{\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3}}) = (3^{2\sqrt{x^{2}-3}+1} - 3^{\sqrt{x^{2}-3} + 1} - 18) \sqrt{x}+ 6x $$ Whati is the maximum value $x$ that fits in the equation?

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Attempt:

$$ \sqrt{x}(9^{\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3}}-6) = (3^{2\sqrt{x^{2}-3}+1} - 3^{\sqrt{x^{2}-3} + 1} - 18) $$ $$ \sqrt{x}(3^{2\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3}}-6) = 3(3^{2\sqrt{x^{2}-3}} - 3^{\sqrt{x^{2}-3}} - 6) $$

Let $y=3^{\sqrt{x^{2}-3}}$, then $ \pm \sqrt{\frac{\ln^{2}(y)}{\ln^{2}(3)} + 3} = x $, and then

$$ \sqrt{x}(y^{2} + y -6) = 3(y^{2} - y - 6) $$ $$ \sqrt{\frac{\ln^{2}(y)}{\ln^{2}(3)} + 3} = x = 9\frac{(y-3)^{2}(y+2)^{2}}{(y+3)^{2}(y-2)^{2}} $$

$$ \sqrt{\frac{\ln^{2}(y)}{\ln^{2}(3)} + 3} = 9\frac{(y-3)^{2}(y+2)^{2}}{(y+3)^{2}(y-2)^{2}} $$

$$ \sqrt{\ln^{2}(y) + 3\ln^{2}(3)} = 9 \ln(3) \frac{(y-3)^{2}(y+2)^{2}}{(y+3)^{2}(y-2)^{2}} $$

After this I have no idea how to continue. Or should we let substitution $X = \sqrt{x^{2}-3}$?

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Actually I received the problem as $$ x(9^{\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3x}}) = (3^{2\sqrt{x^{2}-3}+1} - 3^{\sqrt{x^{2}-3} + 1} - 18) \sqrt{x}+ 6x $$ but I presume that there is a typo in one $\sqrt{x^{2}-3x}$, which I presume that it should be $\sqrt{x^{2}-3}$. IF I was wrong then let us solve the actual problem:

$$ \sqrt{x}(9^{\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3x}}-6) = 3(3^{2\sqrt{x^{2}-3}} - 3^{\sqrt{x^{2}-3}} - 6) $$

$$ \sqrt{x}(9^{\sqrt{x^{2}-3}} + 3^{\sqrt{(x-\frac{3}{2})^{2} - \frac{9}{4}}}-6) = 3(3^{2\sqrt{x^{2}-3}} - 3^{\sqrt{x^{2}-3}} - 6) $$

$$ \frac{\sqrt{x}}{3} = \frac{(9^{\sqrt{x^{2}-3}} - 3^{\sqrt{x^{2}-3}} - 6)}{(9^{\sqrt{x^{2}-3}} + 3^{\sqrt{(x-\frac{3}{2})^{2} - \frac{9}{4}}}-6)} $$

We must have $|x|\ge \sqrt{3}, x\ge 3, x \ge 0$, so $x$ must be $ x \ge 3$. Notice also that the RHS should be positive and it must be less than 1 because $ 3^{\sqrt{(x-\frac{3}{2})^{2} - \frac{9}{4}}} >-3^{\sqrt{x^{2}-3}} $

So we must have $3 \le x < 9$. How to continue from here?

Answer 1

If you're looking for a closed-form solution, you're out of luck with the problem as stated. However, you can fairly easily get an excellent approximation. Let's look at it in this form $$ \frac{\sqrt{x}}{3} = \frac{3^{2\sqrt{x^{2}-3}} - 3^{\sqrt{x^{2}-3}} - 6}{3^{2\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3}}-6} = 1 - \frac{2\cdot3^\sqrt{x^2-3}}{3^{2\sqrt{x^{2}-3}} + 3^{\sqrt{x^{2}-3}}-6} $$ From considering the RHS, we must have $x < 9$. However, because the subtracted term is roughly $2/3^x$ for $x$ near $9$, we know there is a solution slightly less than $9$. Letting $x = 9(1-\epsilon)$ gives $$ \sqrt{1-\epsilon} = 1 - \frac{2\cdot3^\sqrt{81(1-\epsilon)^2-3}}{3^{2\sqrt{81(1-\epsilon)^{2}-3}} + 3^{\sqrt{81(1-\epsilon)^{2}-3}}-6} $$ If we assume that $\epsilon \ll 1$, then the $LHS$ is approximately $1 - \epsilon/2$. The subtracted term on the RHS is already tiny, so we can approximate it by its value at $\epsilon = 0$. This gives $$ \epsilon \approx \frac{4\cdot 3^\sqrt{78}}{9^\sqrt{78}+3^\sqrt{78}-6}\approx\frac{4}{3^\sqrt{78}}. $$ So to first order, $$ x \approx 9 - \frac{4}{3^{\sqrt{78}-2}}\approx 8.9977997 $$ which is within $0.6$ ppm of the numeric value $8.9977945$

Answer 2

Just for the fun of it !

@eyeballfrog provided an elegant, simple and accurate solution.

Starting with the equation he/she used in answer, we could use a Taylor expansion built at $\epsilon=0$ limited to $O(\epsilon^2)$ and ignoring the higher order terms, solving for $0$ the linear equation, get the ugly $$\epsilon=\frac{52\ 3^{\sqrt{78}} \left(-6+3^{\sqrt{78}}+9^{\sqrt{78}}\right)}{13 \left(-6+3^{\sqrt{78}}+9^{\sqrt{78}}\right)^2-2\ 3^{\frac{7}{2}+\sqrt{78}} \sqrt{26} \left(6+9^{\sqrt{78}}\right) \log (3)}\approx 0.0002450663195$$ making $$x=9(1-\epsilon)\approx 8.997794403$$ while the exact solution would be $8.997794531$.