作业帮已知log2^64 +log2^x =5 ,则x的值是?
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已知log2^64 +log2^x =5 ,则x的值是?
已知log2^64 +log2^x =5 ,则x的值是?
已知log2^64 +log2^x =5 ,则x的值是?
log2^64 +log2^x =5
6+log2[x]=5
得 log2[x]=-1
所以
x=1/2
你好，解答如下：
log2^64 = 6
所以log2 x = -1
所以x = 1/2
因为log2^64=6
所以log2^x=-1
所以x=0.5
已知log2^64 +log2^x =5 ,则x的值是?
log2^(x^log2^x)=?
log2(x – 2 ) + 2 = log2(5)log2(x – 2 ) + 2 = log2(5)
log2(x – 2 ) + 2 = log2(5)log2(x – 2 ) + 2 = log2(5)
[log2 1]+[log2 2]+[log2 3]+[log2 4]+[log2 5]+...+[log2 1024]=?[x]表示不超过x的最大整数2为底 答案是8204
解2log2^(x-5)=log2^(x-1)+1
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