logx=5.6, logy=2.4, log(x^2+ y)的值最接近哪個整數?

log x=5.6
x=10^5.6
log y=2.4
y=10^2.4
log(10^5.6^2+10^2.4)
=11.2

Logx=5.6,
logy=2.4,
log(x^2+ y)= log(10^11.2+10^2.4)
又 log(a)在a=10^11.2之斜率為2.74*10^-12
且 log(a)是一嚴格遞增且凹口向下的函數
所以log(10^11.2+10^2.4)＜ log(10^11.2) +(2.31*10^-12)*10^2.4=11.2+2.74*10^-9.6
故取11