# A quick algebra exercise

• Sep 13th 2010, 03:51 AM
klik11
A quick algebra exercise
$\displaystyle 3^{x+1} + 3^{x+2}$

How do I simplify this?
• Sep 13th 2010, 03:57 AM
yeKciM
Quote:

Originally Posted by klik11
$\displaystyle 3^{x+1} + 3^{x+2}$

How do I simplify this?

$\displaystyle 3^{x+1} + 3^{x+2 } = 3 \cdot 3^x + 3^2 \cdot 3^x = 3^x ( 3 + 3^2 ) = 3^x \cdot (3+9 ) = 3^x \cdot 12$
• Sep 13th 2010, 04:00 AM
klik11
I'm Sorry, I forgot to say that the book says that the answer is
$\displaystyle 12*3^{x}$

Could you show me the way to get to that answer?

Edit: I also think you are wrong because it would have been your answer if it there was * and not +.
• Sep 13th 2010, 04:06 AM
yeKciM
Quote:

Originally Posted by klik11
I'm Sorry, I forgot to say that the book says that the answer is
$\displaystyle 12*3^{x}$

Could you show me the way to get to that answer?

thank you ... lol sorry working so many things at the same time .... :D:D:D I edited that one :D up there
• Sep 13th 2010, 04:09 AM
klik11
Could you explain me how come

$\displaystyle 3 \cdot 3^x + 3^2 \cdot 3^x = 3^x ( 3 + 3^2 )$
• Sep 13th 2010, 04:15 AM
yeKciM
Quote:

Originally Posted by klik11
Could you explain me how come

$\displaystyle 3 \cdot 3^x + 3^2 \cdot 3^x = 3^x ( 3 + 3^2 )$

okay ...

$\displaystyle 3^2 = 3^1 \cdot 3^1$

so

$\displaystyle 3^{x+1} = 3^x \cdot 3^1$

now you can write

$\displaystyle 3 \cdot 3^x + 3^2 \cdot 3^x$

here you see both of them have common $\displaystyle 3^x$ so you can pull it out ...

$\displaystyle yx +2x = x (y+2 )$

same thing

$\displaystyle 3 \cdot 3^x + 3^2 \cdot 3^x = 3^x ( 3 + 3^2 )$
• Sep 13th 2010, 05:51 AM
Wilmer
Remember that a^x * a^y = a^(x+y) and (a^x)^y = a^(xy)