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((a^3/2)/(b^3))/((a^-1)/(b^2))

I tried to solve this problem many times, however I tend to get the wrong answer.

Here is the method I tried

(((a^3)^1/2)/(b^3))*... sorry I get confused

i got

(a^2)/(b)

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    $\begingroup$ Please use $\LaTeX$. Does the equation read $$ ((a^3/2)/(b^3))/((a^{-1})/(b^2))$$ or $$((a^{3/2})/(b^3))/((a^{-1})/(b^2))$$? $\endgroup$ – Maximilian Gerhardt Jun 16 '16 at 17:03
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    $\begingroup$ Hello, welcome to math stack exchange. Please format your question using LaTeX. Formatting tips meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Kushal Bhuyan Jun 16 '16 at 17:04
  • $\begingroup$ Note to anyone trying to edit; please don't assume you know the correct interpretation, we need to get that from the original question setter. $\endgroup$ – Joffan Jun 16 '16 at 17:28
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This is a slower explanation. For a fast solution, check the other answer.

Based on your first simplification ($a^{3/2} = (a^{1/2})^{3}$), I take the equation reads

$$((a^{3/2})/(b^3))/((a^{-1})/(b^2))$$ Lets write this out more nicely with fractions, but a division symbol in the middle:

$$\frac{a^{3/2}}{b^3} \div\frac{a^{-1}}{b^2}$$ Instead of dividing by that fraction on the right, we multiply with its inverse. The above expression then reads $$\frac{a^{3/2}}{b^3} \cdot \frac{b^2}{a^{-1}}$$ We multiply numerator and denominator. $$\frac{a^{3/2}\cdot b^2}{b^3\cdot a^{-1}}$$ We can simplify the $b$ expression:$$\frac{b^2}{b^3} =\frac{\color{red}{b^2}}{\color{red}{b^2}\cdot b} = \frac{1}{b}$$ Or, more generally using the law $$\frac{x^a}{x^b} = x^{a-b}$$ Thus, we can also simplify $$\frac{a^{3/2}}{a^{-1}} = a^{3/2 - (-1)} = a^{3/2 + 1} = a^{3/2 + 2/2} = a^{5/2}$$ And the entire expression becomes $$\frac{\color{blue}{a^{3/2}}\cdot \color{red}{b^2}}{\color{red}{b^3}\cdot \color{blue}{a^{-1}}} = \frac{\color{blue}{a^{5/2}}}{\color{red}{b}}$$ Which is the final expression.

For completness: There are lots of ways to get to that answer, e.g. one could simplify $$\frac{a^{3/2}}{a^{-1}}$$ using the law $$a^{-n} = \frac{1}{a^n}$$ which here implies that $$\frac{1}{a^{-1}} = \frac{1}{\frac{1}{a}} = a$$ And then $$x^a\cdot x^b = x^{a+b}$$ to get $$\frac{a^{3/2}}{a^{-1}} = a^{3/2}\cdot a = a^{5/2}$$

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  • $\begingroup$ It was the law that i was stuck with. I have to write it down on my notebook and again thank you so much. $\endgroup$ – SPICA Jun 16 '16 at 17:26
  • $\begingroup$ Im new to this website, so i wonder if there is a "like button" or something so you can earn some benefits. I think it is not enough just to say thank you with words. $\endgroup$ – SPICA Jun 16 '16 at 17:35
  • $\begingroup$ @SPICA You should accept the answer that helped you the most, upvote answers that helped you, and you can also upvote comments. (With your rep you can only do these actions on your own post, you can't upvote just yet however) $\endgroup$ – Colbi Jun 16 '16 at 17:36
  • $\begingroup$ You don't have enough points to upvote yet, but you can mark my answer as "accepted" if that's what you want. You should also take the help tour on math.stackexchange.com/tour $\endgroup$ – Maximilian Gerhardt Jun 16 '16 at 17:36
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If you have:

$$ \frac{\;\;\frac{a^{3/2}}{b^3}\;\;}{\frac{a^{-1}}{b^2}} $$ then it is: $$ \frac{a^{3/2}}{b^3}\frac{b^2}{a^{-1}}=\frac{a^{3/2}\cdot a}{b}=\frac{a^{5/2}}{b} $$

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  • $\begingroup$ Thank you! I understand the law rule but how is $\endgroup$ – SPICA Jun 16 '16 at 17:29
  • $\begingroup$ OOOOOOHHHHHHHHH $\endgroup$ – SPICA Jun 16 '16 at 17:29
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If you have: $$((a^3/2)/(b^3))/((a^{-1})/(b^2))$$ $$=\dfrac{\dfrac{a^3}{2b^3}}{\dfrac{a^{-1}}{b^2}}$$ $$=\dfrac{a^3b^2}{2a^{-1}b^3}$$ $$=\dfrac{a^3}{2a^{-1}b}$$ $$=\dfrac{a^4}{2b}$$

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