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$4^{x-1} + 4^{2-x} = 5 $

I know the result easily but I lack the general reasoning behind it? Should I use $ln$ or other approaches?

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    $\begingroup$ How do you know the result easily? Is it in positive integers? Or all real numbers? Anyway, see this question how to approach such questions. $\endgroup$ – Dietrich Burde Oct 5 '19 at 14:52
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    $\begingroup$ Let $z=4^x$. Then your equation reads $\frac z4+\frac 8z=5$. $\endgroup$ – lulu Oct 5 '19 at 14:55
  • $\begingroup$ Yes, sorry, I forgot to write the solution is in real numbers only. And yes, in this case I see that it should be 4+1 = 5 and I am just trying to find for which x the x-1 = 0 and 2-x = 1 or x-1 = 1 and 2-x = 0. $\endgroup$ – cris14 Oct 5 '19 at 14:56
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You have to turn it into an algebraic equation. This is suggested by $4^x$ and $4^{-x}$. Namely, if you call $4^x=t$ then you obtain an equation of the type $at^2+bt+c=0$.

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Substitute $u=4^{x-1}$ so that ${1\over u}=4^{1-x}$. Then the equation becomes $$u+{4\over u}=5$$ or $$u^2-5u+4=0$$

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