<p>If m and n are positive and 5m^-5 * n^-3 = 20m^3 * n, what is the value of m in terms of n?</p>
<p>(A) 1/4n
(B) 4/n^2
(C) 4/n^3
(D) 2n^2 (Correct)
(E) 4n^2</p>
<p>Can someone kindly explain this in great details. It's pretty confusing.</p>
<p>Well, you’ve gotta get m all by itself. It might help to remember that negative exponents mean the inverse of the positive exponent. So m^(-5) = 1/(m^5).</p>
<p>5m^(-5) * n^(-3) = 20m^3 * n
5/(m^5 * n^3) = 20m^3 * n <– simplify
5/(m^5) * n^4) = 20m^3 <– divide by n
5/(n^4) = 20m^8 <– multiply by m^5
1/(4n^4) = m^8 <– divide by 20
1/(2n^2) = m^4 <– take square root
1/(√(2)n) = m^2 <– take square root again
1/(2^(1/4)√n) = m <– take square root one more time</p>
<p>Annnnd, I should’ve checked that I agreed with your answer before I started typing. The answer you’re giving is equal to m^(-4), not m. Wolfram Alpha agrees with me (see link below). Where’d you get this question?</p>
<p>[5m^(-5)</a>; n^(-3) = 20m^3 * n - Wolfram|Alpha](<a href=“5m^(-5) n^(-3) = 20m^3 * n]5m^(-5) - Wolfram|Alpha”>5m^(-5) n^(-3) = 20m^3 * n - Wolfram|Alpha)</p>
<p>Ooooops, checked the problem, it should be 5m^(5), not -5. Sorry bout tht</p>
<p>As PWN said, you need to get m all by itself.</p>
<p>5<em>m^5</em>n^-3 = 20<em>m^3</em>n</p>
<p>m^5<em>n^-3 = 4 *m^3</em>n (Divide both sides by 5)</p>
<p>m^2<em>n^-3 = 4</em> n (Divide both sides by m^3)</p>
<p>m^2 = 4*n^4 (Multiply both sides by n^3)</p>
<p>m = 2n^2 (take square roots of both sides but take only positive root, since m is positive)</p>
<p>move the m’s to one side and the n’s to the other.</p>
<p>(5m^5)/(20m^3)=n^3 * n</p>
<p>By exponent rules for multiplication and division,</p>
<p>(m^(5-3))/4 = n^(3+1)
(m^2)/4 = n^4
m^2 = 4n^4
m=2n^(4/2)
m=2n^2</p>
<p>5 * m^5 * n^-3 = 20 * m^3 * n</p>
<p>Let n be 2</p>
<p>5 * m^5 * 2^-3 = 20 * m^3 * 2
5 * m^5 * .125 = 40 * m^3
.625 * m^5 = 40 * m^3
m^5 = 64 * m^3
m^2 = 64
m = 8 (m=-2 won’t do since m>0)</p>
<p>The right answer would be one that turns out to be 8 when we plug n=2 into the corresponding expression.
(A) 1/4n
(B) 4/n^2
(C) 4/n^3
(D) 2n^2
(E) 4n^2</p>
<p>Here goes nothing. 
We can quickly rule out A. B. and C as too small.
D produces 2*2^2=8 - that works; E will be greater than D.
That leaves D only.</p>
<p>This question was good practice, thanks! Echoing @PWNtheSAT, where did you get this problem?</p>
<p>Also, @PWNtheSAT, that is like the coolest site ever. Can it do your Calc homework? ;)</p>
<p>Wolfram Alpha is totally bonkers. It can probably help with your calc homework.</p>
<p>
lol:
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<p>Thank you very much guys!</p>
<p>@Risubu, Dr.Chung’s book</p>