exponential equation $$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$
So I have squared both sides and got:
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$
$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$
I don't know what to do now

While this problem succumbs to high-school algebra, with a little college algebra one can go much further to derive recurrences and addition formula for the power sums  $\rm\ s_n\: =\ w^n + w'^n\ $ of the roots of an arbitrary quadratic polynomial $\rm\ f(x)\ =\ (x-w)\ (x-w')\ =\ x^2 - b\ x + c\:.\ $ Namely, we have $\rm\ s_{n+1}\ =\ b\ s_n - c\ s_{n-1}\ $ and, more generally, putting the recurrence in matrix form (e.g. see the Fibonacci case) yields the addition formula $\rm\ s_{m+n} = s_m\ s_n - c\ s_{m-n}\:.\: $ This enables very rapid computation of the sequence by what amounts to repeated squaring of the matrix representing the shift operator (e.g. see the cited Fibonacci case).  For example we obtain a doubling formula via $\rm\:m=n\:$ in the addition formula: $\rm\ s_{\:2\:n}\ =\ s_n^2 - c\ s_0\ =\ s_n^2 - 2\:.\:$ In the example at hand we have $\rm\ s_0,\ s_1,\:\ldots\ =\ 2,10,98,970,9602,95050,940898\ =\ s_6\ $ and, indeed, $\rm\ s_6\ =\ s_3^2 - 2\ =\ 970^2-2\:.\ $ This is a special case of general results about Lucas-Lehmer sequences. For further discussion see Ribenboim: The New Book of Prime Number Records.
