明珠小学对口小区:超越方程 求解

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x+log[2,(2^x-31)]=5

答案x=5

求过程...
log[2,(2^x-31)]
底数2
真数2^x -31

x + log[2,(2^x-31)] = 5

log[2,(2^x-31)] = 5 - x
2^x - 31 = 2^(5 - x)
2^x - 31 = 2^5/2^x

令 u = 2^x
u^2 - 31u - 32 = 0
(u - 32)(u - 1) = 0
u = 32 ................x = 5
u = 1 ..................x = 0 ....... 2^x - 31 < 0