Question

数学题，求详解

已知u≥1，v≥1且(logau)2＋(logav)2＝loga(au2)＋loga(av2)(a>1)，则loga(uv)

Answer 1

解：由题意，
等式右边=loga au² +loga av²=loga a +loga u²+loga a +loga v²
=1+2loga u+1+2loga v
移项得，(loga u-1)²+（loga v-1）²=2²
且原式=loga（uv）=loga u + loga v
令logau=x,logav=y,易知x≥0 ，y≥0.
则（x-1）²+（y-1）²=2²[-π/2 ≤ x ≤ π/2]
且原式=x+y
设x-1=2sina, y-1=2cosa
则x=1+2sina≥o，y=1+2cosa≥0
即：sina≥-1/2，cosa≥-1/2.
则a∈[-π/6,π/2]
则原式=2sina+1+2cosa+1
=2sina+2cosa+2
=2√2sin（a+π/4）+2
且a+π/4∈[π/12,3/4π]
所以，原式=x+y 的最大值为2√2+2; 最小值为2√2×sin（π/12）+2