Question

If $N=\left(2^{lo{g}_{70}^{9800}}\right)\left(5^{lo{g}_{70}^{140}}\right)\left(7^{lo{g}_{70}^2}\right)$ then N is equal to

Answer 1

$N=\left(2^{lo{g}_{70}^{9800}}\right)\left(5^{lo{g}_{70}^{140}}\right)\left(7^{lo{g}_{70}^2}\right)2^{lo{g}_{70}^{9800}}=2^{lo{g}_{70}^{2\times 4900}}=(2{)}^{(lo{g}_{70}^2+lo{g}_{70}^{4900})}=(2{)}^{lo{g}_{70}^2}(2{)}^{lo{g}_{70}^{{70}^2}}=2^{lo{g}_{70}^2}{2}^{2lo{g}_{70}^{70}}=2^{lo{g}_{70}^2}\times 4-(1)$
$(5{)}^{lo{g}_7^{140}0}=(5{)}^{lo{g}_{70}^{2\times 70}}=5^{(lo{g}_{70}^2+lo{g}_{70}^{70})}=5^{lo{g}_{70}^2\times 5^{lo{g}_{70}^{70}}}=5\times 5^{lo{g}_{70}^2}--\left(2\right)$
$N=2^{lo{g}_{70}^{9800}}\times 5^{lo{g}_{70}^{140}}\times 7^{lo{g}_{70}^2}=4\times 2^{lo{g}_{70}^2}\times 5\times 5^{lo{g}^{2}70}\times 7^{lo{g}^{2}70}=20\times (2^{lo{g}_{70}^2}\times 5^{lo{g}_{70}^2}\times 7^{lo{g}_{70}^2})=20(2\times 5\times {)}^{lo{g}_{70}^2}=20(70{)}^{lo{g}_{70}^2}=20\times 2=40(9^{lo{g}_a^b}=b)$