Solve equality with square

$2^{2x}=7\cdot 2^{x+\sqrt{x-1}}+8\cdot 2^{2\sqrt{x-1}}$

$x-1\ge0 \\\sqrt{x-1}=t\ge0\Rightarrow x-1=t^2\Rightarrow x=t^2+1\\2^{2(t^2+1)}=7\cdot2^{t^2+1+t}+2^{3+2t}$
It looks very complicated and I don't know how to move it.
Is there a better way to approach this task?



If $$2^x=a,2^{\sqrt{x-1}}=b$$

Now for real $y,2^y>0$

we have $$0=a^2-7ab-8b^2=(a-8b)(a+b)$$

So, $$2^x=2^{3+\sqrt{x-1}}$$

Set $\sqrt{x-1}=p\ge0\implies x=p^2+1$

Like Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$

as $2\ne\pm1$

$$\implies p^2+1=3+p\iff(p-2)(p+1)=0$$

  • $\begingroup$ thanks for amazing solution! $\endgroup$ – vmahth1 Nov 21 '19 at 0:40

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