a. B > 0. For the equation of X, the logarithm of ax with 10 as the base multiplied by the logarithm of BX with 10 as the base equals - 1. The equation has a solution. Find the range of a / b Thinking,

a. B > 0. For the equation of X, the logarithm of ax with 10 as the base multiplied by the logarithm of BX with 10 as the base equals - 1. The equation has a solution. Find the range of a / b Thinking,

The equation has a solution
( lga+lgx)(lgb+lgx)=-1
Expand lgx square + (LGA + LGB) lgx + lgalgb = - 1
Because the equation has a solution, so △ = (LGA + LGB) square-4 (lgalgb + 1) > 0
Lga-lgb > 2 or lga-lgb100 or 0 〈 A / b

To solve the system of equations x − 2Y = 4 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① 2x + y − 3 = 0 & nbsp; & nbsp; & nbsp; & nbsp; ②

X − 2Y = 4 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① 2x + y − 3 = 0 & nbsp; & nbsp; & nbsp; & nbsp; ②, from ①, x = 2Y + 4, ③ is substituted into ②, 2 (2Y + 4) + Y-3 = 0, the solution is y = - 1, and substituting y = - 1 into ③, x = 2 × (- 1) + 4 = 2, so the solution of the equations is x = 2Y = − 1

To solve the system of equations x − 2Y = 4 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① 2x + y − 3 = 0 & nbsp; & nbsp; & nbsp; & nbsp; ②

X − 2Y = 4 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; ① 2x + y − 3 = 0 & nbsp; & nbsp; & nbsp; & nbsp; ②, from ①, x = 2Y + 4, ③ is substituted into ②, 2 (2Y + 4) + Y-3 = 0, the solution is y = - 1, substituting y = - 1 into ③, x = 2 × (- 1) + 4 = 2