基本的な対数の計算

■問題

次の式を簡単にせよ

${\displaystyle {log}_{\sqrt{3}}8+{log}_{2}3{log}_{9}4+{log}_{\frac{1}{3}}16}$

■答

${\displaystyle 2{log}_{3}2+1}$

■ヒント

対数計算の手順を参照

■解説

${\displaystyle {log}_{\sqrt{3}}8+{log}_{2}3{log}_{9}4+{log}_{\frac{1}{3}}16}$

底の変換公式を用いて，底を$3$に統一する．

${\displaystyle =\frac{{log}_{3}8}{{log}_3\sqrt{3}}+\frac{{log}_{3}3}{{log}_{3}2}\frac{{log}_{3}4}{{log}_{3}9}+\frac{{log}_{3}16}{{log}_3\frac{1}{3}}}$

ここで，対数の性質と指数に関する定義を用いて以下のように式を変形する．

${\displaystyle {log}_{3}8}$${\displaystyle ={log}_3{2}^3}$${\displaystyle =3{log}_{3}2}$

${\displaystyle {log}_3\sqrt{3}}$${\displaystyle ={log}_3{3}^{\frac{1}{2}}}$${\displaystyle =\frac{1}{2}{log}_{3}3}$${\displaystyle =\frac{1}{2}}$

${\displaystyle {log}_{3}3}$${\displaystyle =1}$

${\displaystyle {log}_{3}4}$${\displaystyle ={log}_3{2}^2}$${\displaystyle =2{log}_{3}2}$

${\displaystyle {log}_{3}9}$${\displaystyle ={log}_3{3}^2}$${\displaystyle =2{log}_{3}3}$${\displaystyle =2}$

${\displaystyle {log}_{3}16}$${\displaystyle ={log}_3{2}^4}$${\displaystyle =4{log}_{3}2}$

${\displaystyle {log}_3\frac{1}{3}}$${\displaystyle ={log}_3{3}^{-1}}$${\displaystyle =-{log}_{3}3}$${\displaystyle =-1}$

となる．よって

${\displaystyle =\frac{{log}_3{2}^3}{{log}_3{3}^{\frac{1}{2}}}+\frac{1}{{log}_{3}2}\frac{{log}_3{2}^2}{{log}_3{3}^2}+\frac{{log}_3{2}^4}{{log}_3{3}^{-1}}}$

${\displaystyle =\frac{3{log}_{3}2}{\frac{1}{2}}+\frac{1}{{log}_{3}2}\frac{2{log}_{3}2}{2}+\frac{4{log}_{3}2}{-1}}$

${\displaystyle =6{log}_{3}2+1-4{log}_{3}2}$

${\displaystyle =2{log}_{3}2+1}$

■問題へ戻る

ホーム>>カテゴリー分類>>指数／対数>>対数に関する問題>>基本的な対数の計算

作成：学生スタッフ

最終更新日： 2023年11月29日