$$\left(\frac{5}{5}\right)^n \cdot \left(\frac{9}{10}\right)^n = \left(\frac{1}{5}\right)^n \cdot \left(5 \cdot \frac{9}{10}\right)^n $$

I would like for someone to explain in laymen's terms how to use the following algebraic manipulation function: the left bracket has been multiplied by $5^{-1}$, and the right bracket has been multiplied by $5$. Can someone please explain how this works ?


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    $\begingroup$ Keep in mind that $(a/b)^n=a^n/b^n$. Apply this and it will be clearer. $\endgroup$ – MathematicianByMistake Jan 22 '16 at 21:11
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    $\begingroup$ $$\left(\frac{5}{5}\right)^n*\left(\frac{9}{10}\right)^n=\frac{5^n}{5^n}*\frac{9^n}{10^n}=\frac{1}{5^n}*\frac{5^n*9^n}{10^n}=\left(\frac{1}{5}\right)^n * \left(\frac{5*9}{10}\right)^n$$ $\endgroup$ – Mufasa Jan 22 '16 at 21:11
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    $\begingroup$ If you prefer, you can think of this as a "multiplication by one." You may always multiply by one or add zero and it does not change the value. Here, we have $ab = ab \cdot \frac{c}{c} = (ac)\cdot (\frac{b}{c})$ $\endgroup$ – JMoravitz Jan 22 '16 at 21:15
  • $\begingroup$ This has certainly cleared things up, thank you for your help :) $\endgroup$ – Flewitt Connor Jan 22 '16 at 21:19

$$\begin{align}(5/5)^n \times (9/10)^n &= (5^n/5^n) \times (9/10)^n \\&= 5^n\times(1/5)^n \times (9/10)^n \\&= (1/5)^n \times (5\times9/10)^n\end{align}$$

Because $(a/b)^n=a^n/b^n$ and $a^n\times b^n=(a\times b)^n$.

  • $\begingroup$ Got it ! Thanks very much, I see now that 5/5 = 1 = 5*1/5....and the 5 can be used to multiply the 9 inside the bracket, the 1/5 part remains the same :) $\endgroup$ – Flewitt Connor Jan 22 '16 at 21:18

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