So I have squared both sides and got:



I don't know what to do now


7 Answers 7


You don't have to square the equation in the first place.

Let $y = \sqrt{(5+2\sqrt{6})^x}$, then $\frac{1}{y} = \sqrt{(5-2\sqrt{6})^x}$. Hence you have $y + \frac{1}{y} = 10$ i.e. $y^2 + 1 = 10y$ i.e. $y^2-10y+1 = 0$.

Hence, $(y-5)^2 =24 \Rightarrow y = 5 \pm 2 \sqrt{6}$.

Hence, $$\sqrt{(5+2\sqrt{6})^x} = 5 \pm 2\sqrt{6} \Rightarrow x = \pm 2$$

(If you plug in $x = \pm 2$, you will get $5+2\sqrt{6} + 5-2\sqrt{6} $ which is nothing but $10$)

  • $\begingroup$ How you notice that $\sqrt{(5+2\sqrt{6})^x}=\sqrt{(5-2\sqrt{6})^x}^{-1}$? After checking I know that is true, but I wouldn't do this in the first place $\endgroup$
    – Templar
    Mar 20, 2011 at 19:02
  • $\begingroup$ @Templar: At some stage you will multiply them together (as you did in your first attempt when squaring the left hand side) and find they give $(25-4 \times 6)^{x/2}=1$ $\endgroup$
    – Henry
    Mar 20, 2011 at 19:07
  • $\begingroup$ @Templar: That's what you did when you squared your equation. You get the $1^x$ term from $(\sqrt{5+2 \sqrt{6}} \times \sqrt{5-2 \sqrt{6}}) = 1$. $\endgroup$
    – user17762
    Mar 20, 2011 at 19:16
  • $\begingroup$ Ambikasaran yeah I know but I didn't even think about it... $\endgroup$
    – Templar
    Mar 20, 2011 at 19:19
  • $\begingroup$ If you observe that they are conjugate, you should know from rationalization that their product is an integer. So even if you don't realize is 1, you still know that one is an integer over the other, and this makes the equation easier... $\endgroup$
    – N. S.
    Nov 3, 2011 at 16:34

You've already seen that $(5-2\sqrt{6})(5+2\sqrt{6})=1$ when you squared both sides. This means that $5-2\sqrt{6}=\frac{1}{5+2\sqrt{6}}$, so your last equation can be rewritten as $$\left(\frac{1}{5+2\sqrt6}\right)^x+(5+2\sqrt6)^x+2=100$$ or, letting $y=(5+2\sqrt{6})^x$, $$\frac{1}{y}+y+2=100$$ so $$1+y^2+2y=100y$$ which is quadratic in $y$. Solve this for $y$, then use that solution to solve for $x$.


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.



We have, $5+2\sqrt6 = (\sqrt{2} + \sqrt{3})^2$ and $5 - 2\sqrt6 = (\sqrt3-\sqrt2)^2$

The equation $\Leftrightarrow \left(\sqrt{2} + \sqrt{3}\right)^x + \left(\sqrt3-\sqrt2\right)^x = 10$

Let $$t = \left(\sqrt3+\sqrt2\right)^x = \frac{1}{\left(\sqrt3-\sqrt2\right)^x}$$ with $t\neq 0$

The equation $\Leftrightarrow t + \frac1t - 10 = 0 \Leftrightarrow t^2 - 10t + 1 = 0 $

$\Leftrightarrow t = 5 \pm 2\sqrt6$

With case $t = 5 + 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 + 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5+2\sqrt6) = 2$

With case $t = 5 - 2\sqrt6 \Rightarrow \left(\sqrt3+\sqrt2\right)^x = 5 - 2\sqrt6 \Rightarrow x = \log_{\sqrt3+\sqrt2}(5-2\sqrt6) = -2$.



Observe that $(5 - 2\sqrt{6}) = \frac{1}{5+2\sqrt{6}}$.

So, if we set $a = 5+ 2 \sqrt{6}$, we have \begin{equation} a^x +a^{-x} +2 = 100. \end{equation} The above expression is symmetric in $x$, so if $x = k$ is a solution, $x = -k$ is also a solution. Consider $f(x) = a^x +a^{-x} +2$ for $x>0$. You can show that this is an increasing function for $x > 0$. This implies that there can be at most one value of $x>0$ for which the equality is satisfied. You can check by substitution that $x = 2$ is a solution. So, the only solutions for the equation are $x = +2$ and $x=-2$.


Rather strangely it is possible to spot a solution from the equation simply by realising $$\left(5+2\sqrt6\right)+\left(5-2\sqrt6\right)=10$$ and once you know $5+2\sqrt{6} = \frac{1}{5-2\sqrt{6}}$ then you can spot the other.


Wolfram Alpha can help:


  • $\begingroup$ Enter the following formula into Alpha and look at its use of logarithms to solve the problem: sqrt((5+2*sqrt(6))^x) + sqrt((5-2*sqrt(6))^x) = 10 $\endgroup$
    – Sol
    Mar 20, 2011 at 18:48

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