演習問題

解の存在定理を用いて，方程式 ${\displaystyle 3^x=6x-1}$ が ， ${\displaystyle 0<x<1}$ の区間に実数解を持つことを示せ．

次の式を簡単にせよ．

${\displaystyle -3^2+{\left(-2\right)}^5+8^{\frac{1}{3}}+2^3\cdot 2^2}$

次の式を簡単にせよ．

${\displaystyle -3^2+{\left(-2\right)}^5+8^{\frac{1}{3}}+2^3\cdot 2^2}$

次の式を簡単にせよ．

${\displaystyle -3^2+{\left(-2\right)}^5+8^{\frac{1}{3}}+2^3\cdot 2^2}$

次の式を簡単にせよ．

${\displaystyle -3^2+{\left(-2\right)}^5+8^{\frac{1}{3}}+2^3\cdot 2^2}$

次の式を簡単にせよ．

${\displaystyle \frac{2^3}{9}-{\left(\frac{2}{3}\right)}^2-3^{-2}+{\left(\frac{3}{4}\right)}^{-1}}$

次の式を簡単にせよ．

${\displaystyle \frac{2^3}{9}-{\left(\frac{2}{3}\right)}^2-3^{-2}+{\left(\frac{3}{4}\right)}^{-1}}$

${\displaystyle {log}_{3}27}$

${\displaystyle {log}_{3}27}$

${\displaystyle {log}_{3}27}$

${\displaystyle {log}_{3}27}$

${\displaystyle {log}_2\left(\frac{1}{\sqrt{8}}\right)}$

${\displaystyle {log}_2\left(\frac{1}{\sqrt{8}}\right)}$

${\displaystyle {log}_2\left(\frac{1}{\sqrt{8}}\right)}$

${\displaystyle {log}_2\left(\frac{1}{\sqrt{8}}\right)}$

${\displaystyle {log}_2\left(\frac{1}{\sqrt{8}}\right)}$

${\displaystyle {log}_2\left(\frac{1}{\sqrt{8}}\right)}$

${\displaystyle {log}_2\left(\frac{1}{\sqrt{8}}\right)}$

次の式のグラフを描け．

${\displaystyle y=2^x}$

次の式のグラフを描け．

${\displaystyle y=2^x}$

次の式のグラフを描け．

${\displaystyle y={\left(\frac{1}{2}\right)}^x}$

次の式のグラフを描け．

${\displaystyle y={\left(\frac{1}{2}\right)}^x}$

次の式のグラフを描け．

${\displaystyle y={log}_{2}x}$

次の式のグラフを描け．

${\displaystyle y={log}_{2}x}$

次の式のグラフを描け．

${\displaystyle y={log}_{2}x}$

次の式のグラフを描け．

${\displaystyle y={log}_{2}x}$

次の式のグラフを描け．

${\displaystyle y={log}_{2}x}$

次の式のグラフを描け．

${\displaystyle y={log}_{2}x}$

次の式のグラフを描け．

${\displaystyle y={log}_{\frac{1}{2}}x}$

次の式のグラフを描け．

${\displaystyle y={log}_{\frac{1}{2}}x}$