Originally Posted by

**MarkFL2** Note that:

$\displaystyle 2^x+2\cdot2^x+4\cdot2^x=56$

is not the same as:

$\displaystyle 2^x+(2\cdot2)^x+(4\cdot2)^x=56$

To make things simpler to see, let $\displaystyle u=2^x$ then we have:

$\displaystyle u+2u+4u=56$

Now combine like terms on the left:

$\displaystyle 7u=56$

Divide both sides by 7:

$\displaystyle u=8$

Back-substitute for $\displaystyle u$:

$\displaystyle 2^x=8=2^3$

You see, when I factored in my first post, I used the fact that each term on the left has $\displaystyle 2^x$ as a factor.