已知2^a3^b=2^c3^d=6,证明(a-1)(d-1)=(b-1)(c-1)

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2^a3^b=2^c3^d=6 都同时除以6
=> (2^a3^b)/6=(2^c3^d)/6=1
=> (2^a3^b)/(2*3)=(2^c3^d)/(2*3)=1
=> 2^(a-1)3^(b-1)=2^(c-1)3^(d-1)=1
同时取ln
ln(2^(a-1)3^(b-1))=ln(2^(c-1)3^(d-1))=ln1=0
=>ln(2^(a-1)3^(b-1))=0 =>ln2^(a-1)+ln(3^(b-1))=(a-1)ln2+(b-1)ln3=0
=>(b-1)ln3=-(a-1)ln2 =>ln3/ln2=-(a-1)/(b-1)①

ln(2^(c-1)3^(d-1))=ln1=0
=>=>ln(2^(c-1)3^(d-1))=0 =>ln2^(c-1)+ln(3^(d-1))=(c-1)ln2+(d-1)ln3=0
=>(c-1)ln3=-(d-1)ln2 =>ln3/ln2=-(c-1)/(d-1)②
①=② =>-(a-1)/(b-1)=-(c-1)/(d-1) =>(a-1)(d-1)=(b-1)(c-1)即证

2^a3^b=2^c3^d=6
所以lg2^a3^b=lg2^c3^d=lg6
所以alg2+blg3=lg6
clg2+dlg3=lg6

alg2+blg3=lg6
alg2=lg6-blg3
a=(lg6-blg3)/lg2
a-1=(lg6-blg3)/lg2-1
=(lg6-blg3-lg2)/lg2
=(lg6/2-blg3)/lg2
=(lg3-b(lg3)/lg2
=(1-b)lg3/lg2

clg2+dlg3=lg6
dlg3=lg6-clg2
d=(lg6-clg2)/lg3
d-1=(lg6-clg2)/lg3-1
=(lg6-clg2-lg3)/lg3