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I'm on a chapter about simplifying expression that have indices.

This example has gotten me stumped:

$$\frac{15a^5b^2}{3a^2b^3}= \frac{15}{3} \times \frac{a^5}{a^2} \times \frac{b^2}{b^3} = \frac{5a\frac{1}{3}}{b}$$

I thought the answer would be:

$$\frac{5a^3}{b^{-1}}$$

I should add that I'm coming back to math in life after school, it seems there are things missing from memory.

Thanks.

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    $\begingroup$ no the result is $$\frac{5a^3}{b}$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 5 '18 at 21:22
  • $\begingroup$ @Dr.SonnhardGraubner Thats what he meant with $b^1$ I guess $\endgroup$ – Isham Jan 5 '18 at 21:23
  • $\begingroup$ yes instead of $b$ one Can also write $b^1$ $\endgroup$ – Dr. Sonnhard Graubner Jan 5 '18 at 21:25
  • $\begingroup$ He just made a little mistake...Op it's not $b^{-1}$ but $b^1$ or write $5a^3b^{-1}$ $\endgroup$ – Isham Jan 5 '18 at 21:26
  • $\begingroup$ I thought $b^2$/$b^3$ would be $b^-1$? Rule4 mathematics.laerd.com/maths/indices-intro.php $\endgroup$ – Compton Jan 5 '18 at 21:29
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Note that, $x^ m \times x^n= x^{m+n}$. You can read about the Laws of Indices and see their proofs here

Also note that, $\dfrac{1}{x^m}= x^{-m}$

Using these properties: $\frac{15a^5b^2}{3a^2b^3}$ = $\dfrac{15}{3}\times\dfrac{a^5}{a^2}\times \dfrac{b^2}{b^3}$ = $\dfrac{5a^{3}}{b} = 5a^3 b^{-1}$

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  • $\begingroup$ This is how I thought it should be. $b^2$ / $b^3$ = $b^-1$ It seems others above are cancelling $b^2$ from top and bottom leaving a positive power. $\endgroup$ – Compton Jan 5 '18 at 21:50
  • $\begingroup$ @Compton $x^{-m}=\dfrac{1}{x^m}$... $\endgroup$ – Archer Jan 5 '18 at 21:56
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$$\frac{15a^5b^2}{3a^2b^3} = \left(\frac {15}{3}\right)\left(\frac {a^5}{a^2}\right)\left(\frac {b^2}{b^3}\right)= \frac {5a^3}{b}$$

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