<s>[INST] To solve $2^x+4^x+2^{\lfloor x \rfloor}+4^{\lfloor x \rfloor}+2^{x- \lfloor x \rfloor}-4^{x-\lfloor x \rfloor}=50+\sqrt{50}$ How to solve for positive real $x$: $$2^x+4^x+2^{\lfloor x \rfloor}+4^{\lfloor x \rfloor}+2^{x- \lfloor x \rfloor}-4^{x-\lfloor x \rfloor}=50+\sqrt{50}$$ ? 
 [/INST] Notice the function on the left hand side is increasing (for $x>0$). [And for $x<0$ the left hand side is $\leq6$ so we don't care about it]. The only term that pushes down is $4^{x-[x]}$ but this is $\leq4$.
For $x=2$ we get $4+16+4+16+1-1=48<50+\sqrt{50}$.
For $x=3$ we get $8+64+8+64+1-1>50+\sqrt{50}$.
Therefore the solution is in $(2,3)$. 
The equation becomes 
$$2^x+4^x+4+16+2^{x-2}-4^{x-2}=50+\sqrt{50}$$
Or what is the same
$$a+a^2+4+16+a/4-a^2/16=50+\sqrt{50}$$
where $a=2^x$.
This is a quadratic equation in $a$, so you can solve it.
$a=2^{2+1/2}$ or $a=-4/3-2^{2+1/2}$. The latter cannot be equal to $2^x$.
Then $x=\log_2(2^{2+1/2})=2+1/2$.
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