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I know that $$(577+408\sqrt{2})(577-408\sqrt{2})=577^2-2\cdot408^2=1$$ and I should use this fact to solve find $n$: $$x=\frac{\log n}{\log(577+408\sqrt{2}}$$ where $x$ is the greatest root of the equation $$(577+408\sqrt{2})^x+(577-408\sqrt{2})^x=\frac{226}{15}$$

I tried to apply $\log$ on both sides of the equation above, but I don't know how to proceed due to the fact that there's no property about $\log(A+B)$.

I also applied $\log$ in the first identity, but I got $$\log(577+408\sqrt{2})^x+\log(577-408\sqrt{2})^x=0$$ and not the equation I should solve.

mvfs314
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    I will call $A$ and $B$ those two quadratic irrational numbers in the left hand side of the first equation. You have that $AB=1$ and that $A^x+B^x=\frac{226}{15}$. The second equation gives you that $\log(n)=x\log(A)=\log A^x$. So, $n=A^x$. Now, from $AB=1$ you get that $A^xB^x=1$. So, you have system $C+D=\frac{226}{15}$ and $CD=1$, where $C=A^x$ and $D=B^x$. Solve for $C$. – plop May 07 '21 at 15:35

1 Answers1

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Please note that

$ \displaystyle \small 577 - 408\sqrt{2} = \frac{1}{577+408\sqrt{2}}$

Writing $\displaystyle \small 577 + 408 \sqrt2 = a$

$\displaystyle \small (577+408\sqrt{2})^x+(577-408\sqrt{2})^x=\frac{226}{15}$ becomes,

$\displaystyle \small a^x + \frac{1}{a^x} = \frac{226}{15} = 15 + \frac{1}{15} \implies a^x = \displaystyle \small 15, \frac{1}{15}$

As we have to take greater root, $\small x \log a = \log 15$

So, $\displaystyle \small x=\frac{\log n}{\log(577+408\sqrt{2})} = \frac{\log n}{\log a}$

$\implies \small \log n = \log 15$

Hence $\small n = 15$.

Math Lover
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