Solve $(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$

Question

Find the real values of $x$ which satisfy the equation $(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$

My Attempt $$ e^{10}=e^{(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}}=e^{(5+2\sqrt{6})^{x^2-3}}\cdot e^{(5-2\sqrt{6})^{x^2-3}}=\bigg[e^{(5+2\sqrt{6})}\cdot e^{(5-2\sqrt{6})}\bigg]^{x^2-3}\\ =\Big[e^{10}\Big]^{x^2-3}=e^{10(x^2-3)}\\ \implies x^2-3=1\implies x=\pm2 $$ But my reference gives the solutions $x=\pm2,\pm\sqrt{2}$. Why am I missing the solutions $\pm\sqrt{2}$ in my attempt ?

Note that $(a^b)^c\neq a^{(b^c)}$, so the last equality on your first line is incorrect. — Cheerful Parsnip, Apr 08 '19 at 18:27
One correct way to solve this equation is to call $t= (5+2 \sqrt 6)^{x^2-3}$. Then you get the equation $$t+ \frac{1}{t}=10$$ and proceed from there. — Crostul, Apr 08 '19 at 18:29
Put $a=(5+2\sqrt6)^{x^2-3},b=(5-2\sqrt6)^{x^2-3}$ then $a+b=10$ and $ab=1$ — kingW3, Apr 08 '19 at 18:30
$a^{(b^c)} \ne (a^b)^c$ so you can't say $e^{(a^y)}e^{(b^y)}=(e^ae^b)^y$. It's a coincidence that you got $x=\pm 2$ correctly. — fleablood, Apr 08 '19 at 18:38
https://math.stackexchange.com/questions/202078/solve-sqrt52-sqrt6x-sqrt5-2-sqrt6x-10 — lab bhattacharjee, Apr 08 '19 at 19:00
@fleablood i think thats the reason why I didn't notice that mistake. Thanx. — Sooraj S, Apr 08 '19 at 19:37

score 2 · Accepted Answer · answered Apr 08 '19 at 18:40

2

Your third step is wrong: $$a^{b^x}\neq\left(a^b\right)^x.$$

Let $(5+2\sqrt6)^{x^2-3}=t$.

Thus, $t^2-10t+1=0,$ which gives $t\in\{5+\sqrt{24},5-2\sqrt6\}$.

Thus, $x^2-3=1$ or $x^2-3=-1.$

answered Apr 08 '19 at 18:40

Michael Rozenberg

194,933

score 1 · Answer 2 · answered Apr 08 '19 at 18:31

1

Hint: Note that $$(5+2\sqrt{6})(5-2\sqrt{6})=1$$

answered Apr 08 '19 at 18:31

Dr. Sonnhard Graubner

95,283

Solve $(5+2\sqrt{6})^{x^2-3}+(5-2\sqrt{6})^{x^2-3}=10$

2 Answers2

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