# How can I arrive at $a^6$ for expression $\frac{6a^2*a^0}{2a^{-4}}$?

I am given the expression $$\frac{6a^2*a^0}{2a^{-4}}$$ and am told to simplify. The solution provided is $$a^6$$.

For the numerator I'm aware that $$a^0$$ = 1 which reduces the numerator to just $$6a^2$$.

For the denominator, I'm aware that the negative rule of exponents allows me to rewrite $$2a^{-4}$$ as $$2 * \frac{1}{a^4}$$

But I'm unable to see how to reduce the whole expression to $$a^6$$?

Comments are pointing out that the solution is incorrect. Here's a screen shot of the question and answer in case I've taken the syntax down incorrectly. This is viewable on the free online textbook here under the practice section for exponents and scientific notation.

• Doesn't look right to me; should be $3a^6$. Feb 26 '19 at 17:03
• The net exponent for $a$ is $2+0-(-4)=6$. Add exponents from factors in the numerator, subtract exponents from factors in the denominator.
– MPW
Feb 26 '19 at 17:10

Having arrived at $$\frac{6a^2}{2a^{-4}}$$, multiply by $$a^4$$ on top and bottom to get $$\frac{6a^2×a^4}{2a^{-4}×a^4}=\frac{6a^6}{2a^0}=3a^6$$
$$\frac{6a^2*a^0}{2a^{-4}} = \frac{6a^2*1}{2a^{-4}} = {3a^2}*{a^{-(-4)}} = 3a^6$$
The second transition is due to $$\dfrac{1}{a^{-4}} = a^{-(-4)} = a^4$$.
It is $$3\cdot a^{2-(-4)}$$ and so we get $$3a^{6}$$ and it must be $$a\neq 0$$