Solve the equation exponential radical Solve the equation$$31+8\sqrt{15}=(4+\sqrt{15})^x$$for $x$.
I think you could set up a recursion from the coefficients of $(4+\sqrt{15})^{x}$ to the coefficients of $(4+\sqrt{15})^{x+1}$ then find the general formula using the characteristic equation?
It should look like $a_{n+1}=4a_{n}+15b_{n}$ and $b_{n+1}=4b_{n}+a_{n}$, cancel out one of variables, and then take the characteristic equation.
 A: Note That :$$31+8\sqrt{15}= (4+\sqrt {15})^2$$
So the given equation reduces to 
$$(4+\sqrt {15})^2 = (4+\sqrt {15})^x$$
Equating the powers on both the sides :
$$\boxed{\color{red}{x = 2}}$$
A: We have
$$31+8\sqrt{15}=(4+\sqrt{15})^x \iff x=\frac{\log (31+8\sqrt{15})}{\log (4+\sqrt{15})}$$
A: If $x = \log_{4+\sqrt {15}} (31+8\sqrt {15})$ is not an acceptable solution (and they should specify that it is not; it satisfies all the requirements of a solution; It exists and it is a unique value and it solves) then I'm not really sure there is anything to do but guess.
We can note for any positive integer $k$ that $(4+\sqrt{15})^k = \sum_{j=0}^k (\sqrt{15})^2 \sqrt{15}^j*4^{k-j}C_{k,j}=$
$\sum_{j=0;j\text{ even}}^k 15^{\frac j2}*4^{k-j}C_{k,j} + \sqrt{15}\sum_{j=0;j\text{ odd}}15^{\frac{j-1}2}4^{k-j}C_{k,j}$
So if we can get $31 = 4^{2m} + 15*4^{2m-2}C_{2m\{+1\},2} + .... + \{4;1\}\{2m;1\}15^m$.
And $8 = 4^{2m\pm 1} + 15*4^{2m-3;-1}C_{whatever} + .....$ we'd .... have something.
Now
$31 = 15 + 16 = (\sqrt {15})^2 + 4^2$ and $8\sqrt{15} = 2*4*\sqrt{15}$.
So $31+8\sqrt{15} = 4^2 + 2*4*\sqrt{15} + \sqrt{15}^2 = (4+\sqrt{15})^2$
So $x=\log_{4+\sqrt{15}} (4+\sqrt{15})^2 = 2$.
.....
I don't really like this because it is basically guessing.
But it does seem reasonable that if $x$ is not an integer then $(4+\sqrt {15})^x = \text{a mess}$.  An if $x$ is an integer we must have $31 =$ some power of $4 + $ some power of $15$ + several combinations thereof. Well $8\sqrt{15} =\sqrt{15}($ combination of powers of $4$ and powers of $15$).
And given that $8\sqrt{15} = 2*4\sqrt{15}$ and $31 = 4^2 + 15$ and $x=2$ is a good.
Now, confession.... did I see it right away.  Not really.  I first tried factoring and got $31+ 8\sqrt{15} = 8(4+\sqrt {15}) -1$ which was odd.  Then I figure $31 + 8\sqrt{15} = 31 + 2*4\sqrt{15} = 15 + 16+2*4\sqrt{15} + 16+2*4\sqrt{15} + \sqrt{15}^2$>
A: Using your suggested approach:
$$(4+\sqrt{15})^n=a_n+b_n\sqrt{15}$$
$$a_0=1,b_0=0$$
\begin{align}a_{n+1}+b_{n+1}\sqrt{15}&=(4+\sqrt{15})^{n+1}\\&=(4+\sqrt{15})(a_n+b_n\sqrt{15})\\&=(4a_n+15b_n)+(a_n+4b_n)\sqrt{15}\end{align}
It then suffices to work out a few values:
\begin{array}{c|c|c}n&a_n&b_n\\\hline0&1&0\\1&4&1\\\color{#4477FF}2&\color{#4477FF}{31}&\color{#4477FF}8\end{array}

As an alternative, one could instead numerically approximate using $\sqrt{15}\approx\sqrt{16}=4$ to see that we roughly want to solve
$$31+8\cdot4\approx8^n$$
from which the answer is clear, assuming it must be an integer.
