Properties of exponents: How does numerator of 1/a^{-5/6} become -1 ?

[aaron]

Hi There,

I am trying to relearn maths and am going through Khan academy.

I have gotten to algebra one and mostly I am understanding it however I have hit the below problem and cannot figure it out.

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Simplify the following:

. . . . .\(\displaystyle \dfrac{1}{a^{-\frac{5}{6}}}\)

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Solution:

. . . . .\(\displaystyle \begin{align}\dfrac{1}{a^{-\frac{5}{6}}}\, &=\, \left(a^{-\frac{5}{6}}\right)^{-1}\\
\\
&=\, a^{\frac{5}{6}}\end{align}\)

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I was wondering if anyone could explain the full workings of how the numerator becomes -1 to me please?

Thanks so much for your time.

Regards
Aaron

 

[Dr.Peterson]

Hi There,

I am trying to relearn maths and am going through Khan academy.

I have gotten to algebra one and mostly I am understanding it however I have hit the below problem and cannot figure it out.

---

Simplify the following:

. . . . .\(\displaystyle \dfrac{1}{a^{-\frac{5}{6}}}\)

---

Solution:

. . . . .\(\displaystyle \begin{align}\dfrac{1}{a^{-\frac{5}{6}}}\, &=\, \left(a^{-\frac{5}{6}}\right)^{-1}\\
\\
&=\, a^{\frac{5}{6}}\end{align}\)

---

I was wondering if anyone could explain the full workings of how the numerator becomes -1 to me please?

Thanks so much for your time.

Regards
Aaron

Did you mean the exponent, rather than the numerator?

They are applying the rule that 1/bⁿ = b^-n. I would have directly said, 1/a^-5/6 = a^-(-5/6) = a^5/6, taking b to be a, and n to be -5/6.

What they did is to use the special case 1/b = b^-1, and take b in the rule to be a^-5/6. Then they applied a second rule, (b^m)ⁿ = b^(mn).