log（18）8=a,log（18）5=b
lg8/lg18=a
lg5/lg18=b
lg8/lg5=a/b
3lg2/(1-lg2)=a/b
3blg2=a-alg2
lg2=a/(3b+a)
3lg2/a=lg18
lg18=3/(3b+a)
lg2+lg9=3/(3b+a)
lg9=(3-a)/(3b+a)
所以
log(36)45=lg45/lg36
=(lg5+lg9)/(lg4+lg9)
=(1-lg2+lg9)/(2lg2+lg9)
=(1-a/(3b+a)+(3-a)/(3b+a))/(2a/(3b+a)+(3-a)/(3b+a))
=(3b+a-a+3-a)/(2a+3-a)
=(3b-a+3)/(a+3)这类型的题怎么解？用换底公式变为常用对数再求解