第一题
a=lg(1+1/7) = lg(8/7) = 3lg2 -lg7 (1)
b=lg(1+1/7*7)= lg(50/7^2) = lg50 -2lg7 = lg5 +1 -2lg7 = lg10 -lg2+1-2lg7 =2-lg2-2lg7 (2)
将lg2、lg7当作未知数,解（1）（2）式组成的方程组得
lg2 = (2a-b+2)/7
lg7 = (-a-3b+6)/7
第二题
因 3^a = 4^b =36
取10为底的对数,则
alg3=blg4=lg36
所以
1/a = lg3/lg36
1/b = lg4/lg36
则
2/a+1/b = 2lg3/lg36 + lg4/lg36 = (lg9+lg4)/lg36 = lg36/lg36 = 1