0
$\begingroup$

Are there rules that apply to negative exponents with regard to scientific notation? The specific problem is:

$$\left(6.3\times10^{2}\right)^{-6}$$

I believe the following is correct:

$$\frac{1}{\left(6.3\times10^2\right)^6}$$

However is there a rule that we can apply to the exponent (similar to the rule of multiplying means adding exponents and division means subtracting exponents)? Forgive me as this is a very simple problem, I am back teaching Chem after many years and for some reason this is escaping me.

$\endgroup$
1
  • 1
    $\begingroup$ not sure what your problem is? Yes, $(6.3\times 10^2)^{-6}=\frac{1}{(6.3\times 10^2)^6}$... $\endgroup$ Nov 15, 2015 at 22:24

2 Answers 2

1
$\begingroup$

If I understand the question, then what you wrote is correct. You can verify properties of exponentiation on corresponding Wikipedia page, for example. In particular, \begin{align} \big(\,b^m\,\big)^n &= b^{m\cdot n}& \text{ and }& & b^{-n} &= \dfrac{1}{b^n} \end{align}

$\endgroup$
2
  • 2
    $\begingroup$ This is incorrect: $\left(b^m \right)^n = b^{mn}\neq b^{m+n}$. $\endgroup$
    – mzp
    Nov 15, 2015 at 22:29
  • $\begingroup$ @mzp My bad, thank you for correcting my typo $\endgroup$
    – Vlad
    Nov 15, 2015 at 22:30
0
$\begingroup$

I think the rule you are looking for is that

$$ \left(x^a\right)^b = x^{ab}.$$

Applying it to your case:

$$\left(6.3\times10^{2}\right)^{-6}=\left(6.3^{-6}\times10^{-12}\right)\approx 1.599398 \times 10^{-17}.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .