For a>0, a 1, M,N>0, the product, quotient and power laws are log_a(MN)=log_a M+log_a N, log_a!(MN)=log_a M-log_a N and log_a(Mr)=rlog_a M. Why: let M=au, N=av. Then MN=au+v, MN=au-v, and Mr=aur. A logarithm reads off each exponent. Method: first check arguments and matching bases. To combine, replace multipliers by powers, replace sums by products and differences by quotients. To expand, reverse that order. For p>0, 2ln p-ln3=ln(p2)-ln3=ln!(p23). Exam check: ln(M+N)!= ln M+ln N in general. At M=N=1, the two sides are ln2 and 0, respectively. Keep each original argument positive even after combining.

For \(a>0\), \(a\ne1\), \(M,N>0\), the product, quotient and power laws are \(\log_a(MN)=\log_aM+\log_aN\), \(\log_a\!\left(\dfrac MN\right)=\log_aM-\log_aN\) and \(\log_a(M^r)=r\log_aM\).

Why: let \(M=a^u\), \(N=a^v\). Then \(MN=a^{u+v}\), \(\dfrac MN=a^{u-v}\), and \(M^r=a^{ur}\). A logarithm reads off each exponent.

Method: first check arguments and matching bases. To combine, replace multipliers by powers, replace sums by products and differences by quotients. To expand, reverse that order.

For \(p>0\), \(2\ln p-\ln3=\ln(p^2)-\ln3=\ln\!\left(\dfrac{p^2}{3}\right)\).

Exam check: \(\ln(M+N)\ne\ln M+\ln N\) in general. At \(M=N=1\), the two sides are \(\ln2\) and \(0\), respectively. Keep each original argument positive even after combining.

Why logarithm laws work

Why logarithm laws work