How to Calculate a Logarithm by Hand

You are staring at log₂(50) on a homework problem and wondering how to calculate log without calculator. The calculator is across the room. The exam does not allow one. Here is the method: estimate the value using known powers and interpolation, then refine it with logarithm rules.

Some logarithms are exact. If you can rewrite the argument as a power of the base, the exponent is the answer. For log₁₀(1000), ask: 10 to what power equals 1000? 10³ = 1000, so log₁₀(1000) = 3. For log₂(64), 2⁶ = 64, so log₂(64) = 6. The most common failure is forgetting that the argument must be a perfect power. log₁₀(500) is not exact because 500 is not a power of 10. When the argument is not a direct power, move to the next method.

Memorise the small powers: 2¹ = 2, 2² = 4, 2³ = 8, 2⁴ = 16, 2⁵ = 32, 2⁶ = 64, 2⁷ = 128, 2⁸ = 256, 2⁹ = 512, 2¹⁰ = 1024. For base 10: 10¹ = 10, 10² = 100, 10³ = 1000, 10⁴ = 10,000. For base e: e¹ = e ≈ 2.7183, e² ≈ 7.389, e³ ≈ 20.086. These give you a quick mental check: log₂(50) is between 5 and 6 because 2⁵ = 32 and 2⁶ = 64.

When the argument is not a perfect power, break it into factors. The product rule says log_b(MN) = log_b(M) + log_b(N). The quotient rule says log_b(M/N) = log_b(M) − log_b(N). The power rule says log_b(M^p) = p × log_b(M). These let you rewrite a difficult argument as a combination of numbers whose logarithms you already know.

For example, calculate log₁₀(500). Write 500 = 5 × 100. Then log₁₀(500) = log₁₀(5) + log₁₀(100). You know log₁₀(100) = 2. Now handle log₁₀(5). Since 5 = 10/2, use the quotient rule: log₁₀(5) = log₁₀(10) − log₁₀(2) = 1 − log₁₀(2). If you know log₁₀(2) ≈ 0.3010, then log₁₀(5) ≈ 1 − 0.3010 = 0.6990. Finally, log₁₀(500) ≈ 0.6990 + 2 = 2.6990.

This method works for any argument you can factor into numbers whose logs you have memorised or can estimate. It also works for fractional arguments: log₁₀(0.05) = log₁₀(5/100) = log₁₀(5) − log₁₀(100) = 0.6990 − 2 = −1.3010. The logarithm of a number between 0 and 1 is negative, and that is normal.

You need a small set of constants to four decimal places. Memorise them, write them on your scratch paper, or keep them in your head.

Base 10 (common logarithms):
log₁₀(2) ≈ 0.3010
log₁₀(3) ≈ 0.4771
log₁₀(5) ≈ 0.6990
log₁₀(7) ≈ 0.8451

Base e (natural logarithms):
ln(2) ≈ 0.6931
ln(10) ≈ 2.3026

Base 2 (binary logarithms):
log₂(10) ≈ 3.3219

From these, you can derive many others. For example, log₁₀(6) = log₁₀(2 × 3) = log₁₀(2) + log₁₀(3) = 0.3010 + 0.4771 = 0.7781. log₁₀(8) = log₁₀(2³) = 3 × 0.3010 = 0.9030. The failure case is trying to memorise every value; you only need the primes and the constants above.

When the argument falls between two known powers, gauge the logarithm by interpolation. For log₂(50), you know log₂(32) = 5 and log₂(64) = 6. The argument 50 is roughly 56% of the way from 32 to 64 (50 − 32 = 18, and 64 − 32 = 32, so 18/32 = 0.5625). A linear gauge gives log₂(50) ≈ 5 + 0.5625 = 5.5625. The true value is about 5.6439, so this gauge is slightly low because the logarithm function is concave. Adjust by adding roughly 0.08 to 0.10 for this interval; 5.64 is a better gauge.

For natural logarithms, use ln(2) ≈ 0.6931 and ln(10) ≈ 2.3026. Gauge ln(5): you know ln(4) ≈ 1.3863 (since 2 × 0.6931) and ln(8) ≈ 2.0794 (since 3 × 0.6931). The argument 5 is 25% of the way from 4 to 8, so ln(5) ≈ 1.3863 + 0.25 × (2.0794 − 1.3863) = 1.3863 + 0.1733 = 1.5596. The true value of ln(5) is about 1.6094, so the linear interpolation is off by about 0.05. For a better gauge, use the change of base formula: ln(5) = log₁₀(5) / log₁₀(e). With log₁₀(5) ≈ 0.6990 and log₁₀(e) ≈ 0.4343, you get 0.6990 / 0.4343 ≈ 1.609. This is accurate to three decimal places.

After gauging by hand, verify against a calculator. For log₁₀(500) = 2.6990, a calculator gives 2.69897. The difference is 0.00003, well within rounding error. For log₂(50) ≈ 5.64, a calculator gives 5.6439. The difference is 0.0039, or about 0.07%. The interpolation method is not perfect, but it gives you a value you can trust to two or three decimal places.

The failure case is trusting your gauge without checking the sign. For log₁₀(0.05) you gauged −1.3010. A calculator gives −1.30103. That is close. But if you had forgotten the negative sign, you would be off by 2.602, a 200% error. Always double-check the sign by thinking: is the argument between 0 and 1? If yes, the log is negative for any base greater than 1.

Example 1: log₁₀(2500) Using Known Values

Write 2500 = 25 × 100 = 5² × 100. Then log₁₀(2500) = log₁₀(5²) + log₁₀(100) = 2 × log₁₀(5) + 2. With log₁₀(5) ≈ 0.6990, you get 2 × 0.6990 + 2 = 1.3980 + 2 = 3.3980. Calculator check: log₁₀(2500) = 3.39794. Off by 0.00006.

Example 2: ln(20) Using Natural Log Constants

Write 20 = 2 × 10. Then ln(20) = ln(2) + ln(10) ≈ 0.6931 + 2.3026 = 2.9957. Calculator check: ln(20) = 2.9957. Exact to four decimal places.

Example 3: log₂(100) Using Change of Base

Use the change of base formula: log₂(100) = log₁₀(100) / log₁₀(2). You know log₁₀(100) = 2 and log₁₀(2) ≈ 0.Calculator check: log₂(100) = 6.6439. Off by 0.0001.

Example 4: log₁₀(0.75) Using Quotient Rule

Write 0.75 = 3/4. Then log₁₀(0.75) = log₁₀(3) − log₁₀(4). log₁₀(3) ≈ 0.4771. log₁₀(4) = log₁₀(2²) = 2 × 0.3010 = 0.6020. So log₁₀(0.75) ≈ 0.4771 − 0.6020 = −0.1249. Calculator check: log₁₀(0.75) = −0.1249. Exact to four decimal places.