(1)设log a 2=m,log a 3=n,求a 2m+n 的值;(2)已知10 a =2,10 b =3,求100 2a-b 的值

2025-05-10 21:45:26
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回答1:

(1)∵log a 2=m,∴a m =2.
又log a 3=n,∴a n =3.
于是a 2m+n =a 2m ?a n =2 2 ×3=12.
(2)∵10 a =2,∴a=lg2.
又10 b =3,∴b=lg3.
于是100 2a-b =10 2(lg4-lg3) =(10 lg4 ÷10 lg3 2 = (
4
3
) 2
=
16
9

或100 2a-b =(10 2 2a-b =10 4a-2b =
1 0 4a
1 0 2b
=
(1 0 a ) 4
(1 0 b ) 2
=
2 4
3 2
=
16
9