(1)已知log0.72x<log0.7(x-1),求x的取值范围.
(2)已知lg2=a,lg3=b,试用a,b表示log215.
(1)∵log0.72x<log0.7(x-1),
∴0<x-1<2x,
解得x>1,
故实数x的取值范围是 (1,+∞);
(2)因为lg2=a,lg3=b
所以log215=log230-log22=
-1=lg30 lg2
-1=lg3+1 lg2
.b+1-a a
(1)已知log0.72x<log0.7(x-1),求x的取值范围.
(2)已知lg2=a,lg3=b,试用a,b表示log215.
(1)∵log0.72x<log0.7(x-1),
∴0<x-1<2x,
解得x>1,
故实数x的取值范围是 (1,+∞);
(2)因为lg2=a,lg3=b
所以log215=log230-log22=
-1=lg30 lg2
-1=lg3+1 lg2
.b+1-a a