选C 。
2^a = 5,则 4^a = (2^2)^a = (2^a)^2 = 5^2 = 25;
log8(3) = b,则 8^b = 3,(2^3)^b = 3,2^(3b) = 3;
∴ 4^(3b) = (2^2)^(3b) = [ 2^(3b) ]^2 = 3^
∴ 4^( a - 3b ) = [ 4^a ]/[ 4^(3b
选C 。
2^a = 5,则 4^a = (2^2)^a = (2^a)^2 = 5^2 = 25;
log8(3) = b,则 8^b = 3,(2^3)^b = 3,2^(3b) = 3;
∴ 4^(3b) = (2^2)^(3b) = [ 2^(3b) ]^2 = 3^
∴ 4^( a - 3b ) = [ 4^a ]/[ 4^(3b
这里可以运用指对数互为反函数的性质,先把前项化为a=log2 5, b=三分之一log2 3,再把两者带入问题里就行了