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How to solve for positive real $x$: $$2^x+4^x+2^{\lfloor x \rfloor}+4^{\lfloor x \rfloor}+2^{x- \lfloor x \rfloor}-4^{x-\lfloor x \rfloor}=50+\sqrt{50}$$ ?

drhab
  • 151,093
  • Is $\left[x\right]\leq x$ here? If so then change $\left[x\right]$ into $\lfloor x\rfloor$ by means of \lfloor and \rfloor. – drhab May 20 '14 at 12:31
  • Start by splitting up in cases like $x<1$, $1\le x<2$ et cetera. – drhab May 20 '14 at 12:45

1 Answers1

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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 $x=\log_2(2^{2+1/2})=2+1/2$.