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

distributed the exponted threw.


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"

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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. – Old John Aug 29 '12 at 10:21
you are right,also you can do $8=2^3$ so you take everything with base $2$ – dato datuashvili Aug 29 '12 at 10:23
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? – Old John Aug 29 '12 at 10:41
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 :) – Old John Aug 29 '12 at 10:49
"exponted threw"? Looks OK, but you can simplify $16^{1/2}$. – Gerry Myerson Aug 29 '12 at 13:15
up vote 2 down vote accepted

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: $$4m^{5/4}n^{7/4}$$

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I tend to just keep the fractional exponent, just because it is easy for me. – yiyi Nov 11 '13 at 2:04

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