Re: [中學] 指對數
※ 引述《muta777 (muta)》之銘言:
: 已知 2a+2^a=3
: 2b+2log(b-1)=3 (log是以2為底的)
: 求a+b
令b=2^k+1 則2(2^k+1)+2k=3
-> 2^(k+1)+2(k+1)=3 -> a=k+1
-> b=2^(a-1)+1
a+b=a+2^(a-1)+1=3/2+1=5/2
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