# how to generally approach these types of problems? $4^{x-1} + 4^{2-x} = 5$

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

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

• 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. – Dietrich Burde Oct 5 '19 at 14:52
• Let $z=4^x$. Then your equation reads $\frac z4+\frac 8z=5$. – lulu Oct 5 '19 at 14:55
• 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. – cris14 Oct 5 '19 at 14:56

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 $$at^2+bt+c=0$$.

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