understanding of $2^{-i} \cdot 3^{-i}=\left(\frac16\right)^i$

Question

I'm having trouble understanding how $2^{-i} \cdot 3^{-i}$ reduces to $\left(\frac16\right)^i$ ? Could anyone assist understanding the rules behind this reduction?

Answer 1

Assuming the $i$ in your question is a real number, here are the facts you need.

$x^0 = 1$ for any real $x > 0$.

$x^{a+b} = x^a \times x^b$ for any real $x > 0$ and reals $a,b$.

$(xy)^a = x^a \times y^a$ for any reals $x,y > 0$ and real $a$.

This are non-trivial facts. After all, exponentiation itself is non-trivial to define (see this post for a sketch of the intuition and explanation behind real exponentiation). Any good textbook will define and prove the above facts, but if it is beyond your current level then just make sure that you know the exact conditions under which the facts hold.

Using these facts, you can see the following:

$2^{-i} 3^{-i} = 6^{-i}$.

$6^{-i} 6^i = 6^0 = 1$ and hence $6^{-i} = \large\frac1{6^i}$.

Combining these deductions gives the desired conclusion.