solve
$\displaystyle 2^{\sqrt{x}}+2^x=20$
i know its 4 but how
You'd suspect a perfect square solution so you'd try trial and error ..... try x = 1, 4, 9 ..... whoops go back a second x = 4 aha!
Alternatively:
Let $\displaystyle 2^{\sqrt{x}} = w$. Then:
$\displaystyle w + w^2 = 20 \Rightarrow w^2 + w - 20 = 0 \Rightarrow (w - 4)(w + 5) = 0 \Rightarrow w = 4, \, -5$.
Therefore:
$\displaystyle 2^{\sqrt{x}} = 4 \Rightarrow 2^{\sqrt{x}} = 2^2 \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$.
$\displaystyle 2^{\sqrt{x}} = -5$ has no real solution.
Hi,
Is x an integer ?
I can try to make it but only if x is an integer.
Let's factorize :
$\displaystyle 2^{\sqrt{x}} + 2^x = 2^{\sqrt{x}} ( 1+2^{x-\sqrt{x}}) = 20$
20 is 2²*5
$\displaystyle 1+2^{x-\sqrt{x}}$ is not an even number
So $\displaystyle 2^{\sqrt{x}}$ is 2²
-> $\displaystyle \sqrt{x} = 2$
And you can conclude ^^
Similar to what mr fantastic has done:
$\displaystyle 2^{\sqrt{x}} + 2^{x} = 20$
$\displaystyle 2^{x^{\frac{1}{2}}} + 2^{x} = 20$
$\displaystyle \left(2^{x}\right)^{\frac{1}{2}} + 2^{x} = 20$
Let $\displaystyle w = 2^{x}$
$\displaystyle w^{\frac{1}{2}} + w = 20$
$\displaystyle w^{\frac{1}{2}} = 20 - w$
$\displaystyle w = (20 - w)^{2}$
$\displaystyle w = 400 - 40w + w^{2}$
$\displaystyle 0 = w^{2} - 41w + 400$
$\displaystyle 0 = (w - 16)(w - 25)$
$\displaystyle w = 16 \quad \quad w = 25$
$\displaystyle 2^{x} = 16 \quad \quad 2^{x} = 25$
Solve for x and ignore any erroneous answers.
Let $\displaystyle \mathrm{f}(x)=2^{\sqrt{x}}+2^x-20=\mathrm{e}^{\sqrt{x}\ln{2}}+\mathrm{e}^{x\ln{2} }-20$.
Then $\displaystyle \mathrm{f}'(x)=\mathrm{e}^{\sqrt{x}\ln{2}}\cdot\fr ac{\ln{2}}{2\sqrt{x}}+\mathrm{e}^{x\ln{2}}\cdot\ln {2}>0$ for all $\displaystyle x>0$.
Hence f is a strictly increasing function for all positive x; therefore it cannot cross the x-axis more than once. Since you have found a solution $\displaystyle x=4$, you can be sure there can’t be any more.