Answer:
Option (d) is correct.
for n = 1 [tex]\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216[/tex] holds.
Step-by-step explanation:
Given expression [tex]\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216[/tex]
We have to find the value of n for which the given expression [tex]\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216[/tex]
Consider the given expression [tex]\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216[/tex]
Apply exponent rule [tex]\frac{1}{a^b}=a^{-b}[/tex]
[tex]\frac{1}{\left(\frac{1}{36}\right)^{3n}}=\left(\frac{1}{36}\right)^{-3n}[/tex]
We get,
[tex]216^{n-2}\left(\frac{1}{36}\right)^{-3n}=216[/tex]
Convert [tex]216^{n-2}[/tex] to base 6, we have [tex]216^{n-2}=\left(6^3\right)^{n-2}[/tex]
Thus, the expression becomes,
[tex]\left(6^3\right)^{n-2}\left(6^{-2}\right)^{-3n}=216[/tex]
Apply exponent rule , [tex]\left(a^b\right)^c=a^{bc}[/tex]
We get,
[tex]6^{3\left(n-2\right)}\cdot \:6^{-2\left(-3n\right)}=6^{3\left(n-2\right)-2\left(-3n\right)}[/tex]
[tex]6^{3\left(n-2\right)-2\left(-3n\right)}=216[/tex]
[tex]\mathrm{If\:}a^{f\left(x\right)}=a^{g\left(x\right)}\mathrm{,\:then\:}f\left(x\right)=g\left(x\right)[/tex]
[tex]3\left(n-2\right)-2\left(-3n\right)=3[/tex]
Simplify for n , we have,
n = 1
Thus for n = 1 [tex]\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216[/tex] holds.
