Solve $\sqrt{3}^n=3^3$ I got answer but i could not understand, can anyone explain me ?
$\sqrt{3}^n=3^3$
$\implies{3}^{n/2}=3^3$
$\implies \frac{n}{2}=3\implies n=6$
How we get ${3}^{n/2}$ in the second line and how we get $\frac{n}{2}$ in third line.
 A: Maybe this will make it easier for you to understand:
$(\sqrt3)^n=(3)^3\iff$
$(\sqrt3)^n=(\sqrt3^2)^3\iff$
$(\sqrt3)^n=(\sqrt3)^{2\cdot3}\iff$
$(\sqrt3)^n=(\sqrt3)^6\iff$
$n=6$
A: $ {\sqrt{3}}^{n} = 3^3$
$\Rightarrow$ $({\sqrt{3}}) ^{n} = 3^3$
$\Rightarrow$ $({3}^{1/2}) ^{n} = 3^3$
$\Rightarrow$ $({3}) ^{{(1/2)}.{n}} = 3^3$
$\Rightarrow$ $({3}) ^{n/2} = 3^3$
$\Rightarrow$ $n/2 = 3$
$\Rightarrow$ $n = 6$
A: $\sqrt{3}\equiv3^{\frac{1}{2}}$
The radical--or square root sign--notation is equivalent to the fractional exponent notation. Since raising a rational power to another rational power is equivalent to multiplying the powers together:
$\left(3^{\frac{1}{2}}\right)^n=3^{\frac{n}{2}}$
It appears that you understand how to solve the rest of the problem, or at least don't have any questions on how to do so.
A: Taking logs,  $n.1/2$log$_{3}$ $3$ = $3$ log$_{3}$$3$
Now, log$_{3}$$3$ = $1$.
Hence we have $n/2$ = $3$ $=>$ $n=$6
A: $$\sqrt{x} = x^{\frac{1}{2}}$$
Since if $\sqrt{x}\sqrt{x} = x$ by definition then it follows that $x^kx^k = x^1$ where $2k = 1$ so $k = \frac{1}{2}$. If you choose to agree with this, then
$$\sqrt{3^n} = \sqrt{3}^\frac{n}{2} = 3^3  \implies \frac{n}{2} = 3 \implies n = 6$$
