为何有这些律
对数的每一条律都是指数律换了顶帽子。回想指数积律：b^m · b^n = b^(m+n)——同底幂相乘，指数*相加*。既然对数*就是*指数，这便告诉我们：积的对数应等于各对数之*和*。一边的乘法变成另一边的加法。这种交换——把难运算变成易运算——正是对数被发明的全部缘由。
Proof of the product law (it's just exponents): Let M = b^x so log_b(M) = x Let N = b^y so log_b(N) = y Then M·N = b^x · b^y = b^(x+y) <- exponent product rule So log_b(M·N) = x + y = log_b(M) + log_b(N). That's the whole story: multiply inside, add the logs outside.
三条律
- [[product-law-of-logarithms|积律]]： log_b(M·N) = log_b M + log_b N。积的对数拆成对数之和。（指数积律的镜像。）
- [[quotient-law-of-logarithms|商律]]： log_b(M/N) = log_b M − log_b N。商的对数拆成对数之差。（指数商律的镜像。）
- [[power-law-of-logarithms|幂律]]： log_b(M^p) = p · log_b M。对数内的幂可移到前面作乘数。（指数幂律的镜像。）这一条是解方程的主力。
Using the laws to expand and simplify: Expand log( x^3 · y / z ) = log(x^3) + log(y) - log(z) (product & quotient) = 3·log(x) + log(y) - log(z) (power law on x^3) Condense 2·ln(a) - ln(b) = ln(a^2) - ln(b) (power law) = ln( a^2 / b ) (quotient law) Note what is NOT allowed: log(M + N) does NOT equal log M + log N. The laws only touch products, quotients, powers.
换底
你的计算器只有两个对数键：log（底 10）与 ln（底 e）。那么如何求 log₂ 50？换底公式用你拥有的底重写任意底的对数：log_b(x) = log_c(x) / log_c(b)，其中 c 是任何方便的底——通常是 10 或 e。任选其一，答案相同。
Evaluate log_2(50) using base-10 logs:
log_2(50) = log(50) / log(2)
= 1.69897 / 0.30103
= 5.6439...
Check: 2^5.6439 ≈ 50. Good.
With natural logs you'd get the same:
log_2(50) = ln(50) / ln(2)
= 3.91202 / 0.69315
= 5.6439...