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Solve $(4 + \sqrt15)^x + (4 - \sqrt15)^x =62$

I was able to solve this equation by considering $(4 + \sqrt15)^x$ as some $y$. I got the quadratic equation

$y^2-62y+1=0$.

Therefore, $y = 31 \pm 8 \sqrt15 = (4 + \sqrt15)^x$

I am very close to getting the answer but I dont know how to compare both sides and obtain $x$. Should I just use trial and error (because it works when $x = 2$), but I want to know whether there is a more concrete way yof doing this.

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    $\begingroup$ Forgive me if I'm missing something, but isn't it just $\log_{4+\sqrt{15}}\left(31 \pm 8 \sqrt{15} \right)$? The plus will give you 2, and working through to simplify the minus will give you the other solution. $\endgroup$ Feb 23 '20 at 6:43
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One way is to observe that $31 \pm 8\sqrt{15} = (\sqrt{15})^2 \pm (2)(4)\sqrt{15} + 4^2 = (4 \pm \sqrt{15})^2$, a "sort of" complete the square process.

But that's a bit unsatisfactory since you're basically making a guess (albeit a justified one) that the expression is the square of the surd you want.

You might as well have started by "guessing" $x=2$ to begin with in the original equation.

Either way, you still have to figure out that $x=-2$ is also a valid solution based on the fact that $4 + \sqrt{15}$ is the reciprocal of $4-\sqrt{15}$.

A perhaps more satisfying solution is to use logarithms.

You have $(4 + \sqrt{15})^x = 31 \pm 8\sqrt{15}$

Take logarithms of both sides, it doesn't matter which base as long as it's the same on both sides. Natural logs are fine.

$\log (4 + \sqrt{15})^x = \log(31 \pm 8\sqrt{15})$

$x\log (4 + \sqrt{15}) = \log(31 \pm 8\sqrt{15})$

$x = \frac{\log(31 \pm 8\sqrt{15})}{\log(4 + \sqrt{15})} = \pm 2$

And you get both valid solutions immediately.

The only unsatisfying part about this is that you're forced to use a calculator. But no "leaps of insight" are required - the use of logarithms to solve this form of equation is very standard.

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Hint;

$$31+8\sqrt{15}=2\cdot4\sqrt{15}=(4+\sqrt{15})^2$$

$$\sqrt{31+8\sqrt{15}}=4+\sqrt{15}$$

$$\implies\sqrt{31-8\sqrt{15}}=|4-\sqrt{15}|=4-\sqrt{15}$$ as $4-\sqrt{15}=\dfrac1{4+\sqrt{15}}>0$

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Let $(4+\sqrt{15})^x=y$, then the Eq. becomes $$y^2-62 y+1=0 \implies y=(31\pm 8\sqrt{15})=(4 \pm\sqrt{15}) \implies x=-2,2.$$ Because $$(4 +\sqrt{15})(4-\sqrt{15})=1$$

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$$\begin{align}(4+\sqrt{15})^x+(4-\sqrt{15})^x&=(4+\sqrt{15})^x+(4+\sqrt{15})^{-x}\\ &=e^{x\ln(4+\sqrt{15})}+e^{-x\ln(4+\sqrt{15})}=2\cosh\left(x\ln(4+\sqrt{15})\right)\\ &=62\end{align}$$ So $$\begin{align}x\ln(4+\sqrt{15})&=\pm\cosh^{-1}31=\pm\ln\left(31+\sqrt{31^2-1}\right)\\ &=\pm\ln\left(31+8\sqrt{15}\right)=\pm\ln\left(\left(4+\sqrt{15}\right)^2\right)=\pm2\ln\left(4+\sqrt{15}\right)\end{align}$$ So $x=\pm2$.

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A bit late answer but maybe worth mentioning it.

  • Let $a= 4+\sqrt{15} \stackrel{4-\sqrt{15}=\frac 1{4+\sqrt{15}}}{\Rightarrow} f(x) = a^x + a^{-x}$ is even $(f(x) =f(-x))$, hence we only need to consider $x> 0$, since if $x_0$ is a solution iff $-x_0$ is one.
  • $f'(x) = \ln a\left(a^x - a^{-x}\right)\stackrel{a>1}{>}0 \Rightarrow f'$ is strictly increasing on $(0,+\infty)$. Hence, any solution on $(0,+\infty)$ is unique.

The following part might show how such problems are constructed with other $a$'s and other exponents, as well:

  • Using Vieta you see that $t_1 = a$ and $t_2 = a^{-1}$ are the solutions of $t^2-8t+1 = 0$. Hence,
  • $r_n = a^n + \frac 1{a^n}$ is the solution to the linear recurrence

$$r_{n+2}=8r_{n+1}-r_n \text{ with } r_0 = a^0+\frac 1{a^0}=2, r_1=a+\frac 1a=8$$

$$r_2 = 8\cdot 8-2 = 62 \stackrel{x>0}{\Rightarrow} \boxed{x=2} \text{ is the unique positive solution.}$$

And, because of $f(-x) = f(x)$, the other corresponding unique negative solution is $\boxed{x=-2}$.

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