150 Followers Question

If a = $25^{12}$ , b = $16^{14}$ and c = $11^{16}$ , for to discover the largest, you must to find the logarithm of each number. Then a = log $25^{12}$ , b = log $16^{14}$ and c = log $11^{16}$ . Thence, a = 16.77, b = 16.85 and c = 16.66.

So, a = $10^{16.77}$ , b = $10^{16.85}$ and c = $10^{16.66}$ . Therefore the largest number is b = $16^{14}$ .

$a=25^{12}=5^{24},b=16^{14}=2^{56},c=11^{16}\\ \begin{aligned} a-b&=5^{24}-2^{56}\\ &=(5^{12}-2^{28})(5^{12}+2^{28})\\ &=(5^6-2^{14})(5^6+2^{14})(5^{12}+2^{28})\\ &=(5^3-2^7)(5^3+2^7)(5^6+2^{14})(5^{12}+2^{28})\end{aligned}$ Now notice that $5^3-2^7=-3<0$ ,so $a-b<0\implies b>a$ .Similarly: $\begin{aligned} a-c&=5^{24}-11^{16}\\ &=(5^{12}-11^8)(5^{12}+11^8)\\ &=(5^6-11^4)(5^6+11^4)(5^{12}+11^8)\\ &=(5^3-11^2)(5^3+11^2)(5^6+11^4)(5^{12}+11^8) \end{aligned}$ As $5^3-11^2=4>0$ so $a-c>0\implies a>c$ .Therefore, $b>a>c$ hence $\boxed{b}$ is the largest.