lm(y~x)

lm<-lm(y=x)
summary(lm)

因为你的模型不是线性模型，可以试试用nls来做非线性的拟合

x=c(1,2,3,4,5,6,7,8,9,10,11,12)
y1 =c(7098.00,7498.00,7848.00,8254.00,8761.00,8801.12,8951.32,9325.03,9680.90,10200.00,11000.00,12360.74)
fit1 <- nls(y1 ~ a * b^x, start = list(a=1, b = 2))
plot(y1 ~ x)
lines(seq(1, 12, by = 0.1), predict(fit1, data.frame(x=seq(1, 12, by = 0.1))))

y=a*x^b
取对数 得 log(y)=b*log(ax)
log(y)=b*log(a)+b*log(x)
转化为线性
所以, 用可得b*log(a)和b

x和y1的长度不一样啊