SaitechAI • Applications of Logarithms (Multiplication, Division, Powers, Roots)

Quick Reference

Rules

\[ \begin{aligned} \log_b(MN) &= \log_b M + \log_b N,\\[2pt] \log_b\!\left(\frac{M}{N}\right) &= \log_b M - \log_b N,\\[2pt] \log_b(M^k) &= k\,\log_b M,\\[2pt] \log_b\!\left(\sqrt[n]{M}\right) &= \tfrac{1}{n}\,\log_b M,\\[2pt] \log_b M &= \dfrac{\log_k M}{\log_k b}. \end{aligned} \]

Domain & Notes

\(b>0,\; b\neq 1,\; M>0,\; N>0\).
Antilog (base \(b\)): \(\operatorname{antilog}_b(x)=b^{\,x}\).
Rounding arises from tabulated/approximate logs.

Multiplication via Logarithms

To compute \(A\times B\): add logs and take antilog.

Example 1: \(24 \times 35\)

Factorize: \(24=3\cdot 2^3,\; 35=5\cdot 7\). Then \[ \begin{aligned} \log(24\cdot 35) &= \log 3 + 3\log 2 + \log 5 + \log 7\\ &\approx 0.4771 + 3(0.3010) + 0.6990 + 0.8451\\ &= 0.4771 + 0.9030 + 0.6990 + 0.8451\\ &= 2.9242. \end{aligned} \] Antilog: \[ 24\cdot 35 \approx 10^{2.9242}=10^{0.9242}\times 10^{2} \approx 8.4\times 100 = 840. \] Exact value: \(24\times 35=840\) (agreement up to rounding of logs).

Example 2: \(2.5 \times 40\)

\[ \log(2.5\cdot 40) = \log 2.5 + \log 40 = (\log 5 - \log 2) + (\log 4 + \log 10). \] Using \(\log 5=0.6990,\; \log 2=0.3010,\; \log 4=0.6021,\; \log 10=1\): \[ \log(2.5\cdot 40) \approx (0.6990-0.3010)+(0.6021+1)=0.3980+1.6021=2.0001\approx 2. \] Antilog: \(2.5\cdot 40 \approx 10^{2}=100\) (exact: \(100\)).

Division via Logarithms

To compute \(A\div B\): subtract logs and take antilog.

Example 1: \(\dfrac{840}{35}\)

\[ \log\!\left(\frac{840}{35}\right)=\log 840 - \log 35. \] Note \(840=84\cdot 10\) with \(84=2^2\cdot 3\cdot 7\): \[ \log 840 = (\!2\log 2 + \log 3 + \log 7) + \log 10 \approx (2\cdot 0.3010 + 0.4771 + 0.8451) + 1 = 1.9242 + 1 = 2.9242. \] And \(\log 35=\log 5 + \log 7 \approx 0.6990+0.8451=1.5441\). Hence \[ \log\!\left(\frac{840}{35}\right) \approx 2.9242 - 1.5441 = 1.3801, \] so \[ \frac{840}{35} \approx 10^{1.3801}=10^{0.3801}\times 10 \approx 2.4 \times 10 = 24. \] Exact value: \(24\).

Example 2: \(\dfrac{980}{14}\)

\(980=98\cdot 10,\; 98=2\cdot 7^2\). \[ \log 980 = (\log 2 + 2\log 7) + \log 10 \approx (0.3010 + 2\cdot 0.8451) + 1 = 2.9912. \] \[ \log 14 = \log 2 + \log 7 \approx 0.3010 + 0.8451 = 1.1461. \] \[ \log\!\left(\frac{980}{14}\right) \approx 2.9912 - 1.1461 = 1.8451 \;\Rightarrow\; \frac{980}{14} \approx 10^{1.8451} = 10^{0.8451}\times 10 \approx 7\times 10 = 70. \] Exact value: \(70\).

Powers via Logarithms

To compute \(M^k\): use \(\log(M^k)=k\,\log M\), then antilog.

Example 1: \(2.5^3\)

\[ \log(2.5^3)=3\log 2.5 = 3(\log 5 - \log 2) \approx 3(0.6990-0.3010) = 3(0.3980)=1.1940. \] Antilog: \[ 2.5^3 \approx 10^{1.1940} \approx 15.6 \quad (\text{exact } 15.625). \]

Example 2: \(7^{2.5}\)

\[ \log(7^{2.5}) = 2.5\,\log 7 \approx 2.5 \times 0.8451 = 2.1128. \] Antilog: \[ 7^{2.5} \approx 10^{2.1128}=10^{0.1128}\times 10^2 \approx 1.295 \times 100 \approx 129.5. \] (Direct calculator gives \(\approx 129.645\); rounding of \(\log 7\) causes the small difference.)

Roots via Logarithms

To compute \(\sqrt[n]{M}\): use \(\log(\sqrt[n]{M})=\frac{1}{n}\log M\), then antilog.

Example 1: \(\sqrt{5}\)

\[ \log(\sqrt{5})=\tfrac{1}{2}\log 5 \approx \tfrac{1}{2}\times 0.6990 = 0.3495 \;\Rightarrow\; \sqrt{5} \approx 10^{0.3495} \approx 2.237. \] (Exact \(\sqrt{5}\approx 2.23607\).)

Example 2: \(\sqrt[3]{15.625}\)

Note \(15.625=\dfrac{125}{8}\). \[ \log 15.625 = \log 125 - \log 8 = 3\log 5 - 3\log 2 \approx 3(0.6990) - 3(0.3010) = 1.1940. \] Then \[ \log\!\big(\sqrt[3]{15.625}\big)=\tfrac{1}{3}\times 1.1940=0.3980 \;\Rightarrow\; \sqrt[3]{15.625} \approx 10^{0.3980} \approx 2.5. \] (Exact value: \(2.5\).)

Common Log Values (Base 10, 4 d.p.)

\(x\)	\(\log_{10} x\)	\(x\)	\(\log_{10} x\)
2	0.3010	6	0.7781
3	0.4771	7	0.8451
4	0.6021	8	0.9031
5	0.6990	9	0.9542
10	1.0000	2.5	0.3980

Applications of Logarithms: Multiplication, Division, Powers, Roots