基本的な対数の計算

■問題

次の式を簡単にせよ

${\displaystyle \frac{{log}_{2}9+{log}_{2}3}{3}+\frac{2{log}_{2}5+{log}_{2}125}{1-{log}_{2}10}}$

■答

${\displaystyle {log}_{2}3-5}$

■ヒント

対数計算の手順を参照

■解説

${\displaystyle \frac{{log}_{2}9+{log}_{2}3}{3}+\frac{2{log}_{2}5+{log}_{2}125}{1-{log}_{2}10}}$

ここで，与式の${\displaystyle {log}_{2}9+{log}_{2}3}$ と${\displaystyle {log}_{2}25+{log}_{2}125}$ ,${\displaystyle 1-{log}_{2}10}$ を 対数計算の基本を用いて式を簡単にする．

${\displaystyle {log}_{2}9+{log}_{2}3}$${\displaystyle ={log}_2{3}^2+{log}_{2}3}$${\displaystyle =2{log}_{2}3+{log}_{2}3}$${\displaystyle =3{log}_{2}3}$

${\displaystyle {log}_{2}25+{log}_{2}125}$${\displaystyle ={log}_2{5}^2+{log}_2{5}^3}$${\displaystyle ={log}_2{5}^5}$${\displaystyle =5{log}_{2}5}$

${\displaystyle 1-{log}_{2}10}$${\displaystyle =1-\left({log}_{2}2+{log}_{2}5\right)}$${\displaystyle =1-1-{log}_{2}5}$${\displaystyle =-{log}_{2}5}$

よって

${\displaystyle =\frac{3{log}_{2}3}{3}+\frac{5{log}_{2}5}{-{log}_{2}5}}$

${\displaystyle ={log}_{2}3-5}$

■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