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$$13^{2x+1} - 13^x -12=0$$

This is equality that i didn't understand because there is $+1$ on power. If there wasn't +1 on power it would be same as quadratic equality

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    $\begingroup$ Hint: $13^{2x+1} - 13^x -12=13y^2 - y -12=0$ $\endgroup$ – C.F.G Oct 23 '17 at 8:29
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    $\begingroup$ $t=13^x$ gives $13t^2-t-12=0$ . Only the positive solution $t=1$ gives a solution, namely $x=0$. $\endgroup$ – Peter Oct 23 '17 at 8:30
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    $\begingroup$ $13^{2x + 1} = 13^{2x} \times 13$ $\endgroup$ – S.H.W Oct 23 '17 at 8:31
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It'a basic index law: $a^{m+n}= a^m\cdot a^n$

Now, your equations simplifies to:

$13^{2x}\cdot13- 13^x-12=0$

Let $13^x= k$.

Thus, $13k^2-k-12=0 \implies k=-\dfrac{12}{13}\quad \mathrm{or} \quad k= 1$
$\implies13^x=-\frac{12}{13} $ or $13^x=1$

Hence, the only real solution is $x=0$

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  • $\begingroup$ Hm I still don't get it "Thus $13k^2-k-12=0$ which yields $13^x=-\frac{12}{13} $ or $13^x=1$" - how did you get rid of the $13k^2$? And why $13^x=1$? $\endgroup$ – Max Oct 23 '17 at 9:28
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Hint: Let $t=13^x$ and solve $13t^2-t-12=0$.

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  • $\begingroup$ where is +1 $13^{2x+1}$ $\endgroup$ – user494654 Oct 23 '17 at 8:33
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    $\begingroup$ As I told you $13^{2x + 1} = 13^{2x} \times 13$ $\endgroup$ – S.H.W Oct 23 '17 at 8:34
  • $\begingroup$ $13t^2=13\times 13^{2x}=13^{2x+1}$. $\endgroup$ – C.F.G Oct 23 '17 at 8:34
  • $\begingroup$ Understood @C.F.G $\endgroup$ – user494654 Oct 23 '17 at 8:37
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You can get rid of the $+1$ by using the fact that $a^{b+c}=a^b\cdot a^c$ and writing $$13^{2x+1} = 13^{2x}\cdot 13^1$$

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