Given that $\sqrt{3}\times27^{n}=1$, find the value of n.

1. Express each term as a power of 3.

$\sqrt{3}=3^{\frac12}$ and $27=3^3$.

So
$\sqrt{3}\times 27^n=3^{\frac12}\times (3^3)^n.$

2. Use laws of indices to combine powers of the same base.

$(3^3)^n=3^{3n}$

Hence
$3^{\frac12}\times 3^{3n}=3^{\frac12+3n}.$

3. Equate the exponent to zero and solve for $n$.

Since $1=3^0$,
$3^{\frac12+3n}=3^0.$

Therefore,
$\frac12+3n=0$
$3n=-\frac12$
$n=-\frac16.$