Question 1174281
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log rules:
(1) log(ab) = log(a)+log(b)
(2) log(a^n) = n*log(a)<br>
1/2 log(base 3) of M = log(base 3) of M^(1/2)  (rule (1))
3log(base 3) of N = log(base 3) of N^3  (rule (1))<br>
log(base 3) of M^(1/2) + log(base 3) of N^3 = log(base 3) of ((M^1/2)(N^3))  (rule 2)<br>
Now the equation is<br>
log(base 3) of ((M^1/2)(N^3)) = 1<br>
Translating that to exponential form gives us<br>
(M^1/2)(N^3) = 3^1 = 3<br>
Solving for M in terms of N:<br>
{M)(N^6) = 9  (squaring both sides)
M = 9/N^6<br>