Question

2^a2^b=2^c2^d=100, 求证: (a+1)(d+1)=(b+1)(c+1).

Answer 1

问：2^a2^b=2^c2^d=100, 求证: (a+1)(d+1)=(b+1)(c+1).

问：有点小急

答：过程稍稍有点复杂

答：第一步a+b = log2(100)c+d = log2(100)将式子变形，得到：a = log2(100) - bc = log2(100) - d

答：然后将a和c带入证明式中，有：(a+1)(d+1)=(b+1)(c+1)= (log2(100) - b + 1)(d + 1) = (b + 1)(log2(100) - d + 1)

答：然后展开得到log2(100)d - bd + log2(100) - b + d + 1 = bd - b + log2(100)d - d + log2(100) + 1化简可得：b + d = log2(100) - 1

答：然后将b+d的值带入前面的式子，可以得到：(a+1)(d+1)=(b+1)(c+1)= (log2(100) - b + 1)(d + 1) = (b + 1)(log2(100) - d + 1)= (log2(100) - b + 1)(log2(100) - d + 2)/(log2(100) - 1) = (b + 1)(c + 2)/(log2(100) - 1)

答：化简后可得：(a+1)(d+1)=(b+1)(c+1)= (log2(100) - b + 1)(log2(100) - d + 2) = (b + 1)(c + 2)将b+d=log2(100)-1带入，化简后可得：(a+1)(d+1)=(b+1)(c+1)= (log2(100) - b + 1)(log2(100) - log2(100) + 1) = (b + 1)(log2(100) - b + 1)= (log2(100) - b + 1)^2 = (b + 1)(log2(100) - b + 1)

答：(b + 1)(b - log2(100) + 3) = 0因此，b = log2(100) - 3，代回原式可得：a = 5 - b = 2c = 5 - d = 3

答：最后证明(a+1)(d+1)=(b+1)(c+1)= (2+1)(4+1) = (3+1)(3+1)

问：b➕d那部没太懂

答：哪里？

问：然后展开得到log2(100)d - bd + log2(100) - b + d + 1 = bd - b + log2(100)d - d + log2(100) + 1化简可得：b + d = log2(100) - 1把c代入不应该是 log2（100）b嘛

答：不是，是把那个式子拆开然后再代入b+d

答：你看第一步就已经化简出来了第一步a+b = log2(100)c+d = log2(100)将式子变形，得到：a = log2(100) - bc = log2(100) - d

问：还是没懂 第三步不应该是这样吗log2(100)d - bd + log2(100) - b + d + 1 = log2（100）b-bd+b+log2（100）-d+1

答：不是

答：亲亲是这样的，你把右边那个式子用（b+1）(c+1)代入

答：(a+1)(d+1)=(b+1)(c+1)左边的代入就是log2(100)d - bd + log2(100) - b + d + 1 ，右边就等于bd - b + log2(100)d - d + log2(100) + 1

答：你先别管bd了

答：先代ac就可以了

问：代c 右边肯定是log2（100）b啊咋是log2（100）d

答：c=log2(100)-d