Question

已知lg2=a lg3=b 求log(4)12+log(2)15=

Answer 1

log(4)12+log(2)15=lg12/lg4+lg15/lg2

=lg2*2*3/lg2*2+lg(10*3/2)/lg2

=（a+a+b）/（a+a）+(1 + b - a)/a

=1+b/2a+1/a+b/a-1

=（b+2+2b）/a

=（3b+2）/a

2.log(8)9=lg3*3/lg2.*2*2

=（b+b）/（a+a+a）

=2b/3a

3.log(3)32=lg32/lg3

=lg2*2*2*2*2/lg3

=（a+a+a+a+a）/b

5a/