若正实数a,b满足log2(a)+log2(3b)=0则,3^a+27^b的最小值

题目

若正实数a,b满足log2(a)+log2(3b)=0则,3^a+27^b的最小值
对数的式子化简得到3ab=1.3^a+27^b≥2√3^3ab=2√3.

答案

3^a+27^b
=3^a+3^(3b)
≥2√[3^a*3^(3b)]
=2√[3^(a+3b)]
≥2√3^(2√(a*3b))
=2√3^(2*1)
=2√3^2
=2*3
=6
你把3^a*3^(3b)乘错了