Solve for x...

**Hellbent** · November 17th 2010, 11:25 AM

Hi,

Solve for x:

$(a.)\,4^x - 13(2^x) + 40 = 0$

This is supposed to have two solutions (quadratic), but I've only progressed two steps. I believe these two steps to be incorrect. Argh!

$4^x - 26^x + 40 = 0$
$-22^x + 40 = 0$

**pickslides** · November 17th 2010, 11:36 AM

They are incorrect.

$13(2^x)\neq 26^x$ and $4^x - 26^x \neq -22^x$

Recall BODMAS and 'adding like terms' for reasons why.

Try this

$\displaystyle 4^x - 13(2^x) + 40 = 0$

$\displaystyle (2^2)^x - 13(2^x) + 40 = 0$

$\displaystyle (2)^{2x} - 13(2^x) + 40 = 0$

$\displaystyle (2^x)^{2} - 13(2^x) + 40 = 0$

Now making $2^x = a$ you get

$\displaystyle a^{2} - 13a + 40 = 0$

Can you continue? What do you get?

**Hellbent** · November 17th 2010, 12:18 PM

Originally Posted by pickslides

Can you continue? What do you get?

Thanks.

$(a - 5)(a - 8) = 0$
$a = 5; a = 8$

$2^x\,=\,5; 2^x\,=\,8$

$\log(2^x) = \log\,5;\,\,\,\,\,\,\,\,\,\log\,(2^x) = \log\,8$

$x = \frac{\log\,5}{\log\,2};\,\,\,\,\,x = \frac{\log\,8}{\log\,2}$

$x \approx 2.32;\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 3$

$13(2^x)\neq 26^x\,\,\,\,\,\,\,\,\,\,\,4^x - 26^x \neq -22^x$
'Like' here means: of the same base?

**pickslides** · November 17th 2010, 04:49 PM

Originally Posted by Hellbent

$13(2^x)\neq 26^x\,\,\,\,\,\,\,\,\,\,\,4^x - 26^x \neq -22^x$
'Like' here means: of the same base?

Yep, like this

$4^x - (5)4^x = -(4)4^x$

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