Answer:
n = 1
Step-by-step explanation:
We need to solve this equation for "n".
We first have to recognize the denominator and numerator as a "same" base number.
We know that 216 and 36 can be written as powers of 6. So, we write:
[tex]\frac{(6^3)^{n-2}}{(\frac{1}{6^2})^{3n}}=216[/tex]
Now, we can write the denominator using the rule:
[tex]a^b=\frac{1}{a^{-b}}[/tex]
So, it becomes:
[tex]\frac{(6^3)^{n-2}}{(\frac{1}{6^2})^{3n}}=216\\\frac{(6^3)^{n-2}}{(6^{-2})^{3n}}=216[/tex]
Now, we can use the rule:
[tex](a^z)^b=a^{zb}[/tex]
So, we have:
[tex]\frac{(6^3)^{n-2}}{(6^{-2})^{3n}}=216\\=\frac{6^{3n-6}}{6^{-6n}}=216[/tex]
When we have same base, we can write it together using the identity:
[tex]\frac{a^x}{a^y}=a^{x-y}[/tex]
Thus,
[tex]6^{(3n-6)-(-6n)}=216[/tex]
Writing RHS as 6^3 and solving, we have:
[tex]6^{3n-6+6n}=6^3\\6^{9n-6}=6^3\\9n-6=3\\9n=9\\n=1[/tex]
Thus,
n = 1
