$\displaystyle 2x ^ {2+x}= \frac{1}{4^{x+1} } $

Printable View

- Sep 23rd 2008, 06:21 AMhana_102need help simplyfying
$\displaystyle 2x ^ {2+x}= \frac{1}{4^{x+1} } $

- Sep 23rd 2008, 06:47 AMMoo
- Sep 23rd 2008, 06:59 AMhana_102
- Sep 23rd 2008, 08:10 AMSoroban
Hello, hana_102!

I'm 99.99% sure there is a typo . . .

Quote:

$\displaystyle 2^{2+x}\;=\;\frac{1}{4^{x+1}} $

We have: .$\displaystyle \left(2^{x+2}\right)\left(4^{x+1}\right) \;=\;1$

. . . . . . . $\displaystyle \left(2^{x+2}\right)\left(2^2\right)^{x+1} \;=\;1$

. . . . . . . $\displaystyle \left(2^{x+2}\right)\left(2^{2x+2}\right) \;=\;1$

. . . . . . . . . . . . $\displaystyle 2^{3x+4} \;=\;1$

. . . . . . . . . . . . $\displaystyle 2^{3x+4} \;=\;2^0$

. . . . . . . . . . . $\displaystyle 3x + 4 \;=\;0$

. . . . . . . . . . . . . .$\displaystyle \boxed{x \;=\;\text{-}\frac{4}{3}}$

- Sep 23rd 2008, 08:14 AMhana_102
- Sep 23rd 2008, 08:20 AMMoo