How do I solve this exponential-surdic equation? I have been trying to solve a problem today, but have had very little to no progress. The problem is:
Find the real values of $x$ satisfying $$(26 + 15\sqrt{3})^x + 6(2 + \sqrt{3})^x + (2 - \sqrt{3})^x - 5(7 + 4\sqrt{3})^x = 5$$
Clear explanations would surely be appreciated.
 A: Let $r=2+\sqrt3$, then
$$r^2=7+4\sqrt3$$
$$r^3=26+15\sqrt3$$
$$r^{-1}=2-\sqrt3$$
Thus the equation becomes
$$r^{3x}+6r^x+r^{-x}-5r^{2x}=5$$
which is a quartic polynomial in $r^x=s$:
$$s^3+6s+1/s-5s^2=5$$
$$s^4-5s^3+6s^2-5s+1=0$$
$$(s^2-4s+1)(s^2-s+1)=0$$
The real roots of this polynomial (coming from the left factor) are $2+\sqrt3$ and $2-\sqrt3$. Thus the solutions to the original equation are $x=\pm1$.
A: Notice, that $26+15\sqrt3=(2+\sqrt3)^3$ and $7+4\sqrt3=(2+\sqrt3)^2$. Now the equation reads $$(2 + \sqrt{3})^{3x} + 6(2 + \sqrt{3})^x + (2 - \sqrt{3})^x - 5(2 + \sqrt{3})^{2x} = 5$$
Let's multiply both parts of it by a nonzero $(2 + \sqrt{3})^x$. We receive $$(2 + \sqrt{3})^{4x} + 6(2 + \sqrt{3})^{2x} + ((2 - \sqrt{3})(2 + \sqrt{3}))^x - 5(2 + \sqrt{3})^{3x} = 5(2 + \sqrt{3})^x$$
The term $((2 - \sqrt{3})(2 + \sqrt{3}))^x$ equals 1. Let's introduce $a=(2+\sqrt3)^x$. Then we get $$a^4-5a^3+6a^2-5a+1=0$$ This is $$(a^2-4a+1)(a^2-a+1)=0$$ Roots are $2-\sqrt3$ and $2+\sqrt3$, which means that $x=1$ is the solution and another one is $\log_{2+\sqrt3}(2-\sqrt3)$
