4
$\begingroup$

The problem as given: $(4m^3n)^\frac{1}{4} \times (8mn^3)^\frac{1}{2}$

distributed the exponted threw.

$(4^{1/4}m^{3/4}n^{1/4})(8^{1/2}m^{1/2}n^{3/2})$

Then because $a^na^m = a^{n+m} \; \& \; a^{n}b^{n} = (ab)^n$

$(2^2)^{1/4} (8^{1/2})(m^{3/4 + 1/2})( n^{1/4+6/4}) = 16^{1/2}(m^{5/4})n^{7/4}$

am I correct?

edited to fix mistake pointed out by "Old John"

  • 2
    Almost right - at one point you seem to have replaced $3/2$ with $3/4$ instead of $6/4$, but I think that is the only error.2012-08-29
  • 0
    you are right,also you can do $8=2^3$ so you take everything with base $2$2012-08-29
  • 0
    Very nearly there! in the line where you have $n^{1/4 + 3/4}$, should that not be $n^{1/4 + 6/4}$, giving $n^{7/4}$ in the final answer?2012-08-29
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    A possibly useful tip for these things: If you are unsure of an answer, try evaluating the original expression and your answer with some values for $m, n$ using a calculator. If they agree you *might be right*, but if they disagree, you *must* be wrong :)2012-08-29
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    "exponted threw"? Looks OK, but you can simplify $16^{1/2}$.2012-08-29
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    $16^{1/2} = \sqrt{16}$2013-09-21

1 Answers 1

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As pointed out in the comments, your answer looks good. The one thing I would change is that I would replace $16^{1/2}$ with $\sqrt{16}=4$, giving you: $$4m^{5/4}n^{7/4}$$

  • 0
    I tend to just keep the fractional exponent, just because it is easy for me.2013-11-11