段考公式整理

定義域

l o g_{a} (r) = R

a^{R} = r

a^{l o g_{a} (r)} = r

a^{l o g_{a} (r)} \times a^{l o g_{a} (s)} = a^{l o g_{a} (r) + l o g_{a} (s)}

a^{l o g_{a} (r)} \times a^{l o g_{a} (s)} = r s

1.式和2.式同取

l o g_{a}

l o g_{a} (a^{l o g_{a} (r) + l o g_{a} (s)}) = l o g_{a} (r s)

l o g_{a} (r) + l o g_{a} (s) = l o g_{a} (r s)

由

~~$l o g_{a} (r) + l o g_{a} (s) = l o g_{a} (r s)$ 同理可證~~

a^{l o g_{a} (r)} \div a^{l o g_{a} (s)} = a^{l o g_{a} (r) - l o g_{a} (s)}

a^{l o g_{a} (r)} \div a^{l o g_{a} (s)} = r \div s

1.式和2.式同取

l o g_{a}

l o g_{a} (a^{l o g_{a} (r) - l o g_{a} (s)}) = l o g_{a} (r \div s)

l o g_{a} (r) - l o g_{a} (s) = l o g_{a} (r \div s)

l o g_{a} (r^{t}) = l o g_{a} (r \times r \times r \times \dots \times r)

由

l o g_{a} (r) + l o g_{a} (s) = l o g_{a} (r s)

可得

l o g_{a} (r \times r \times r \times \dots \times r)

= l o g_{a} (r) + l o g_{a} (r) + l o g_{a} (r) + \dots + l o g_{a} (r)

= t \times l o g_{a} (r)

a^{l o g_{a} (r)} = r

l o g_{b} (a^{l o g_{a} (r)}) = l o g_{b} (r)

使用脫帽子公式

l o g_{b} (a^{l o g_{a} (r)}) = l o g_{a} (r) \times l o g_{b} (a)

l o g_{b} (r) = l o g_{a} (r) \times l o g_{b} (a)

l o g_{b} (r) \div l o g_{b} (a) = l o g_{a} (r)

換底公式

l o g_{(a^{t})} (r) = l o g_{a} (r) \div l o g_{a} (a^{t}) = l o g_{a} (r) \div t

也是換底