$\displaystyle 2^{2x+1}=2^{2x}\times2^{1}=4^{x2}$ Is that right?

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- Oct 14th 2008, 11:45 AMFlexibleIndicies
$\displaystyle 2^{2x+1}=2^{2x}\times2^{1}=4^{x2}$ Is that right?

- Oct 14th 2008, 12:02 PMmasters
No, that is not correct.

$\displaystyle 2^{2x+1}=2^{2x} \cdot 2^1$ is true. This cannot be simplified further.

$\displaystyle 2^{2x} \cdot 2 \neq 4^{2x}$

Remember the rule for multiplying same bases:

$\displaystyle x^a \cdot x^b=x^{a+b}$

Example: $\displaystyle 2 \cdot 2^2 \neq 4^2$

$\displaystyle 2 \cdot 2^2 = 2^3 = 8$ - Oct 14th 2008, 12:13 PMicemanfan
However, you can write $\displaystyle 2^{2x+1}$ as $\displaystyle 2 \cdot 4^x$.