求（log4 3+log8 3）*（log3 2+log9 2）。（前一个数为底数，后一个为真数）

用换底公式
原式=(lg3/lg4+lg3/lg8)(lg2/lg3+lg2/lg9)
=(lg3/2lg2+lg3/3lg2)(lg2/lg3+lg2/2lg3)
=[(1/2+1/3)(lg3/lg2)][(1+1/2)(lg2/lg3)]
=(1/2+1/3)(1+1/2)
=5/4

(log3 2+log9 2)*(log4 3+log8 3)
=(lg2/lg3+lg2/lg9)*(lg3/lg4+lg3/lg8)
=(lg2/lg3+lg2/2lg3)*(lg3/2lg2+lg3/3lg2
=3/2*lg2/lg3*5/6*lg3/lg2
=3/2*5/6=5/4

（log4 3+log8 3）*（log3 2+log9 2）
=(lg3/2lg2+lg3/3lg2)(lg3/lg2+lg2/2lg3)
=(1/2+1/3)(1+1/2)
=5/4

log43+log83=1/2log2 3+1/3log2 3=5/6log2 3
log3 2+log 92=log3 2+1/2log3 2=3/2log3 2
相乘=5/4