设f(x)=a的a-2/1次方,f(lga)=根号10 求a的值

已知f（x）=A^(x-1/2),f(lgA)=√10
因f（x）=A^(x-1/2),
所以f(lgA)=A^(lgA-1/2)=√10
A^(lgA-1/2)=√10=10^(1/2)
lgA-1/2=logA[10^(1/2)]
=(1/2)logA(10)
=1/(2lgA)
2(lgA)^2-lgA-1=0
(2lgA+1)(lgA-1)=0
lgA=1或lgA=-1/2
A=10或A=10^(-1/2)=√10 /10