1. Solve for x...

Hi,

Solve for x:

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

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

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

2. They are incorrect.

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

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

Try this

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

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

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

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

Now making $\displaystyle 2^x = a$ you get

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

Can you continue? What do you get?

3. Originally Posted by pickslides
Can you continue? What do you get?
Thanks.

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

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

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

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

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

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

4. Originally Posted by Hellbent

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

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