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The question says solve for $x$: $(\frac{1+\sqrt{5}}{2})^{2012}$+$(\frac{1+\sqrt{5}}{2})^{2013}$=$(\frac{1+\sqrt{5}}{2})^{x}$. You were allowed to use a calculator, but you couldn't just plug it in because the calculator can't calculate that many digits. Originally I plugged $\frac{1+\sqrt{5}}{2}$ into my calculator and got about $1.6$. I then rewrote the problem as $1.6^{2012}+1.6^{2013}=1.6^{x}$. I took the natural logarithm of both sides and got. $2012\ln(1.6)+2013\ln(1.6)=x\ln(1.6)$. I then solved for $x$ and got $4025$, which is the wrong answer. The correct answer is $2014$. I managed to find another way to do it and get this answer, but I was wondering what was wrong with the way I did it originally? As a side note, in case you were wondering I did solve it by writing it as: $1.6^{2012}(1+1.6)=1.6^{x}$. $1+1.6=2.6=1.6^2$. Therefore $1.6^{2012}\times1.6^2=1.6^x$, and $x$ has to be $2014$.

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    Simply because $ln(a + b) \neq ln(a) + ln(b)$. Be careful when you take the logarithm of both sides. – 2017-02-22

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When you take log of both sides it should be as follows:

$ln(1.6^{2012}+1.6^{2013})=xln(1.6)$

You can't separate those logarithms on the LHS as you did.

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divide through by the first term, what remains is $$ 1 + \phi = \phi^{x - 2012}. $$ Since $1 + \phi = \phi^2,$ we need $x - 2012 = 2$ or $x = 2014.$

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    While you're correct in that this is a valid solution to the problem, the OP was asking specifically why their original method didn't work. – 2017-02-22
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Note the special relation $\phi^2 = \phi + 1$. Multiply both sides by $\phi^{2012}$ and...