対数の問題です！

質問

(1) 8^log2(3)
(2) log9√27
(3) 9(log8x)^2-8log4x+4=0
(4) log3(18)^2+3/2log3(3/2)-1/2log3(2)
が分かりません。
教えて頂けたら幸いです

回答No.1

asuncion · ベストアンサー

自分は対数の底を[]で書きます。
(1)
8^(log[2]3) = (2^3)^log[2]3 = 2^(3log[2]3) = 2^(log[2]27) = 27
(2)
log[9]√27 = log[9](3^(3/2)) = (3/2)log[9]3 = (3/2) * (1/2) = 3/4
(3)
9(log[8]x)^2 - 8log[4]x + 4 = 0
真数条件からx > 0
log[8]x = (log[2]x)/3, log[4]x = (log[2]x)/2
log[2]x = tとおくと、与式はt^2 - 4t + 4 = 0
(t - 2)^2 = 0, t = 2
log[2]x = 2よりx = 4, これはx > 0を満たす。
∴x = 4
(4)
log[3]18^2 + (3/2)log[3](3/2) - (1/2)log[3]2
= 2log[3](2*(3^2)) + (3/2)(log[3]3 - log[3]2) - (1/2)log[3]2
= 2(log[3]2 + 2) + (3/2)(1 - log[3]2) - (1/2)log[3]2
= 2log[3]2 + 4 + 3/2 - (3/2)log[3]2 - (1/2)log[3]2
= 11/2