# 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 $x$?

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?

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 $$X = \sqrt{x^{2}-3}$$?

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?

• But this is an equation? Do you want to find the solution of it? – Dr. Sonnhard Graubner May 25 at 14:15
• @Dr.SonnhardGraubner yes but the largest one – Arief Anbiya May 25 at 14:16
• in $(3^{2\sqrt{x^{2}-3}+1} - 3^{\sqrt{x^{2}-3} + 1} - 18) \sqrt{x}+ 6x$ is $\sqrt x +6x$ multiplied by $(3^{2\sqrt{x^{2}-3}+1} - 3^{\sqrt{x^{2}-3} + 1} - 18)$ or just $\sqrt x$ then $6x$ is added, summary: try to use parentheses in the last part. – anas pcpro May 25 at 15:25

## 2 Answers

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 $$8.9977945$$

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 $$8.997794531$$.