I'm confused. Why is it that for a problem in the form of: $(2^{x+1})(2^{x-1})$ we get $2^{2x}$ instead of $4^{2x}$; Shouldn't we multiply the $2$s by each other..?
Similarly for a problem like: $(2^{x+1})(4^{x-1})$, why do we get $2^{3x-1}$ rather than $8^{2x}$?
And for $(3^{2x}/3^{x-1})$ we get $3^{x+1}$...shouldn't the $3$s cancel?
Recall that $$\large a^b\cdot a^c = a^{b+ c}$$
$$\large \dfrac{a^b}{a^c} = a^{b - c}$$ $$\large \left(a^b\right)^c = a^{bc}$$
So, using these "laws of exponents", we have: $$\begin{align} (1)\quad \large 2^{(x + 1)}\cdot 2^{(x - 1)} & = \large 2^{(x + 1) + (x - 1)}\\ \\ & = \large 2^{2x}\end{align}$$ $$ $$ $$\begin{align} (2)\quad \large 2^{x+1} \cdot 4^{x-1} & = \large 2^{x + 1} \cdot \left(2^2\right)^{(x-1)}\\ \\ &= \large \large 2^{x + 1} \cdot 2^{2(x-1)} \\ \\ & = \large 2^{(x+1) + 2(x-1)} \\ \\ & = \large 2^{3x - 1}\end{align}$$
Now, for the last question, I'll let you try to work that one out.
