<s>[INST] How to solve this equation using logarithms? I have to solve for all real values of $x$.
$(5+2\sqrt6)^{x^2-3}+(5-2\sqrt6)^{x^2-3}=10$
I tried to take $\log_{10}$ on both sides but could not do this.
How do I do this?Thanks for any hint or answer!!
 [/INST] Hint.
note that $$(5-2\sqrt{6})=\frac{1}{5+2\sqrt{6}}$$
and use the substitution
$$
\left(5+2\sqrt{6}\right)^{x^2-3x}=y
$$
so the equation becomes:
$$
y+\frac{1}{y}=10
$$
that becomes a second degree equation (multiply by $y \ne 0$). Solve this equation and you can find the final solution (without logarithms). 

$$
y^2-10y+1=0 \quad \Rightarrow \quad y=5\pm \sqrt{25-1}=5\pm 2\sqrt{6}
$$
so you have the two solutions:
$$
\left(5+2\sqrt{6}\right)^{x^2-3x}=5+ 2\sqrt{6} $$
$$\left(5+2\sqrt{6}\right)^{x^2-3x}=5- 2\sqrt{6}=\left(5+ 2\sqrt{6} \right)^{-1}
$$
so:
$$
x^2-3x= 1 \quad \lor \quad x^2-3x= -1
$$
Now,  can you find $x$?
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