# Math Help - solve

1. ## solve

solve
$2^{\sqrt{x}}+2^x=20$

i know its 4 but how

2. Originally Posted by perash
solve
$2^{\sqrt{x}}+2^x=20$

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 $2^{\sqrt{x}} = w$. Then:

$w + w^2 = 20 \Rightarrow w^2 + w - 20 = 0 \Rightarrow (w - 4)(w + 5) = 0 \Rightarrow w = 4, \, -5$.

Therefore:

$2^{\sqrt{x}} = 4 \Rightarrow 2^{\sqrt{x}} = 2^2 \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$.

$2^{\sqrt{x}} = -5$ has no real solution.

3. Originally Posted by mr fantastic
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 $2^{\sqrt{x}} = w$. Then:

$w + w^2 = 20 \Rightarrow w^2 + w - 20 = 0 \Rightarrow (w - 4)(w + 5) = 0 \Rightarrow w = 4, \, -5$.

Therefore:

$2^{\sqrt{x}} = 4 \Rightarrow 2^{\sqrt{x}} = 2^2 \Rightarrow \sqrt{x} = 2 \Rightarrow x = 4$.

$2^{\sqrt{x}} = -5$ has no real solution.
thank you mr fantastic
but if
$2^{\sqrt{x}} = w$
then
$2^{2\sqrt{x}} = w^2$
not

$2^{{x}} = w^2$

4. Hi,

Is x an integer ?

I can try to make it but only if x is an integer.

Let's factorize :

$2^{\sqrt{x}} + 2^x = 2^{\sqrt{x}} ( 1+2^{x-\sqrt{x}}) = 20$

20 is 2²*5
$1+2^{x-\sqrt{x}}$ is not an even number

So $2^{\sqrt{x}}$ is 2²

-> $\sqrt{x} = 2$

And you can conclude ^^

5. Originally Posted by perash
thank you mr fantastic
but if
$2^{\sqrt{x}} = w$
then
$2^{2\sqrt{x}} = w^2$
not

$2^{{x}} = w^2$
Whoops!

6. Similar to what mr fantastic has done:

$2^{\sqrt{x}} + 2^{x} = 20$
$2^{x^{\frac{1}{2}}} + 2^{x} = 20$
$\left(2^{x}\right)^{\frac{1}{2}} + 2^{x} = 20$

Let $w = 2^{x}$

$w^{\frac{1}{2}} + w = 20$
$w^{\frac{1}{2}} = 20 - w$
$w = (20 - w)^{2}$
$w = 400 - 40w + w^{2}$
$0 = w^{2} - 41w + 400$
$0 = (w - 16)(w - 25)$
$w = 16 \quad \quad w = 25$
$2^{x} = 16 \quad \quad 2^{x} = 25$

Solve for x and ignore any erroneous answers.

7. Originally Posted by o_O
Similar to what mr fantastic has done:

$2^{\sqrt{x}} + 2^{x} = 20$
$2^{x^{\frac{1}{2}}} + 2^{x} = 20$
$\left(2^{x}\right)^{\frac{1}{2}} + 2^{x} = 20$

Mr F sadly adds: Unfortunately, $(2^{x})^{1/2} = 2^{x/2}$, not $2^{x^{1/2}}$ ......

Let $w = 2^{x}$

$w^{\frac{1}{2}} + w = 20$
$w^{\frac{1}{2}} = 20 - w$
$w = (20 - w)^{2}$
$w = 400 - 40w + w^{2}$
$0 = w^{2} - 41w + 400$
$0 = (w - 16)(w - 25)$
$w = 16 \quad \quad w = 25$
$2^{x} = 16 \quad \quad 2^{x} = 25$

Solve for x and ignore any erroneous answers.
Unfortunately that doesn't work either! But thanks for making me look less of a bonehead than I was!

8. Let $\mathrm{f}(x)=2^{\sqrt{x}}+2^x-20=\mathrm{e}^{\sqrt{x}\ln{2}}+\mathrm{e}^{x\ln{2} }-20$.

Then $\mathrm{f}'(x)=\mathrm{e}^{\sqrt{x}\ln{2}}\cdot\fr ac{\ln{2}}{2\sqrt{x}}+\mathrm{e}^{x\ln{2}}\cdot\ln {2}>0$ for all $x>0$.

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 $x=4$, you can be sure there can’t be any more.