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In a school, on the first day 1 student was absent, on the second day 2 students were absent, on the third day 4 students were absent, on the fourth day 8 students were absent. If same pattern continued till 6^th day, how many students would be absent on the 6^th day?

Find the value of :-
${\left(2\right)}^3\times {\left(3\right)}^3$

Simplify: $\frac{6^0+7^0+8^0}{6^0\times 7^0\times 8^0}$

${\mathrm{a}}^6\times {\mathrm{a}}^5\times {\mathrm{a}}^0=\mathrm{a}$^_____

What will come in place of $\left(?\right)$ in the following question?

${\left(\frac{1}{729}\right)}^{\frac{2}{3}}\times {\left(\frac{1}{1296}\right)}^{\frac{1}{4}}÷{\left(\frac{1}{2187}\right)}^{\frac{2}{7}}=?$

The following can be written in exponential form:

$\mathrm{b}\times \mathrm{b}\times \mathrm{b}\times \dots \dots \dots \dots \dots 12$ times$=b^k$. Find the value of $k$.

If the exponential form of $6\times \mathrm{x}\times \mathrm{x}\times \mathrm{x}\times \mathrm{z}\times \mathrm{z}\times \mathrm{z}$ is in the form $6{\mathrm{x}}^k{\mathrm{z}}^m$, then find the value of $k+m$.

If $(2000{)}^{12}=4.096\times {10}^k$, then the value of $k$is: