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

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.

• 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. Feb 23 '20 at 6:43

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.

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 $$4-\sqrt{15}=\dfrac1{4+\sqrt{15}}>0$$

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 $$(4 +\sqrt{15})(4-\sqrt{15})=1$$

\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 $$x=\pm2$$.

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 $$\boxed{x=-2}$$.