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I have the following exponential equation: $3\sqrt 5\cdot 2^n=(1+\sqrt 5)^n-(1-\sqrt 5)^n$. How do I solve it?

P.S. I can already see $n=4$, but I am interested in how to solve this using logarithms.

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  • $\begingroup$ The addition makes it not so straight forward to use logarithms to solve this. I don't even know how to solve equations of the form $a^n+b^n=c$ in general. $\endgroup$ – Arthur Dec 18 '18 at 19:06
  • $\begingroup$ It may help to note that $$a^n=e^{n\ln a}$$ so then you could at least get everything in the same base... $\endgroup$ – clathratus Dec 18 '18 at 19:10
  • $\begingroup$ @clathratus thanks I will try that now $\endgroup$ – DarkAtom Dec 18 '18 at 19:12
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\begin{align} 3\sqrt5\,2^n&=(1+\sqrt 5)^n-(1-\sqrt 5)^n \tag{1}\label{1} \end{align}

Let's denote $1+\sqrt 5=a$, $\sqrt 5-1=b,\ b>0$.

Note that \begin{align} a\cdot b&=4=2^2 . \end{align}

Consider $n=2m$, $m\in\mathbb{Z}$.

Then \eqref{1} transforms into \begin{align} 3\sqrt5\,2^{2m}&=a^{2m}-(-b)^{2m} \tag{2}\label{2} ,\\ 3\sqrt5\,(2^2)^m&=a^{2m}-((-b)^2)^m ,\\ 3\sqrt5\,(ab)^m&=a^{2m}-(b^{2m}) ,\\ 3\sqrt5\,\left(\frac{a}b\right)^m &=\left(\frac{a}b\right)^{2m} -1 . \end{align}

Let $(\tfrac{a}b)^m=x$,

\begin{align} x^2-3\sqrt5 x-1&=0 ,\\ x&=\tfrac72+\tfrac32\sqrt5 \quad\text{(ignore negative root)} ,\\ (\tfrac{a}b)^m&=\tfrac72+\tfrac32\sqrt5 ,\\ m&= \frac{\ln(\tfrac72+\tfrac32\sqrt5 )}{\ln(\tfrac32+\tfrac12\sqrt5)} \\ &= \frac{\ln((\tfrac32+\tfrac12\sqrt5)^2)}{\ln(\tfrac32+\tfrac12\sqrt5)} \\ &= \frac{2\ln(\tfrac32+\tfrac12\sqrt5)}{\phantom{2}\ln(\tfrac32+\tfrac12\sqrt5)} =2 ,\\ n&=2m=4 . \end{align}

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  • $\begingroup$ Ok, this is what I was looking for, but shouldn't the case $n=2m+1$ also be covered? $\endgroup$ – DarkAtom Dec 18 '18 at 20:15
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    $\begingroup$ Why don't you try it yourself, follow similar steps and you will see that $m \notin \mathbb{Z}$ therefore discarding this case. $\endgroup$ – Naweed G. Seldon Dec 18 '18 at 20:31
  • $\begingroup$ @DarkAtom: yes, you should check this case too. $\endgroup$ – g.kov Dec 18 '18 at 20:53
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This equation is just $\frac{1}{\sqrt{5}}\phi^n-\frac{1}{\sqrt{5}}\psi^n=F_n=3$, where $F_n$ denotes the $n$th Fibonacci number, so the answer is $n=4$.

See Binet's Formula

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hint

Observe that

$$\frac{1-\sqrt{5}}{2}=\frac{-2}{1+\sqrt{5}}$$

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