Question

For what value of n does 216^n-2/(1/36)^3n= 216

Answer 1

Answer is D. 1

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Answer 2

Answer:

Option (d) is correct.

for n = 1  $\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216$ holds.

Step-by-step explanation:

Given expression $\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216$

We have to find the value of n for which the given expression $\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216$

Consider the given expression  $\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216$

Apply exponent rule $\frac{1}{a^b}=a^{-b}$

$\frac{1}{\left(\frac{1}{36}\right)^{3n}}=\left(\frac{1}{36}\right)^{-3n}$

We get,

$216^{n-2}\left(\frac{1}{36}\right)^{-3n}=216$

Convert $216^{n-2}$ to base 6, we have $216^{n-2}=\left(6^3\right)^{n-2}$

Thus, the expression becomes,

$\left(6^3\right)^{n-2}\left(6^{-2}\right)^{-3n}=216$

Apply exponent rule , $\left(a^b\right)^c=a^{bc}$

We get,

$6^{3\left(n-2\right)}\cdot \:6^{-2\left(-3n\right)}=6^{3\left(n-2\right)-2\left(-3n\right)}$

$6^{3\left(n-2\right)-2\left(-3n\right)}=216$

$\mathrm{If\:}a^{f\left(x\right)}=a^{g\left(x\right)}\mathrm{,\:then\:}f\left(x\right)=g\left(x\right)$

$3\left(n-2\right)-2\left(-3n\right)=3$

Simplify for n , we have,

n = 1

Thus for n = 1  $\dfrac{216^{n-2}}{(\frac{1}{36})^{3n} }=216$ holds.