Solve for x the following equation:

$${\log _{(2x + 3)}}(6{x^2} + 23x + 21) = 4 - {\log _{(3x + 7)}}(4{x^2} + 12x + 9)\,$$

Find the sum of the series : $$$\sum\limits_{r = 0}^n {{{\left( { - 1} \right)}^r}\,{}^n{C_r}\left[ {{1 \over {{2^r}}} + {{{3^r}} \over {{2^{2r}}}} + {{{7^r}} \over {{2^{3r}}}} + {{{{15}^r}} \over {{2^{4r}}}}..........up\,\,to\,\,m\,\,terms} \right]} $$$

If $$a > 0,\,b > 0$$ and $$\,c > 0,$$ prove that $$\,c > 0,$$ prove that $$\left( {a + b + c} \right)\left( {{1 \over a} + {1 \over b} + {1 \over c}} \right) \ge 9$$

If $$n$$ is a natural number such that
$$n = {p_1}{}^{{\alpha _1}}{p_2}{}^{{\alpha _2}}.{p_3}{}^{{\alpha _3}}........{p_k}{}^{{\alpha _k}}$$ and $${p_1},{p_2},\,\,......,\,{p_k}$$ are distinct primes, then show that $$In$$ $$n \ge k$$ $$in$$ 2