2
6.64385619
3.32192809
4.60517019
2
6.64385619
5
2
6.64385619
3.32192809
4.60517019
2
6.64385619
5
The Change of Base Calculator converts logarithms between different bases using the change of base formula. If you know $$\log_a(x)$$ and need $$\log_b(x)$$, this calculator performs the conversion instantly. It also provides the value in all common bases ($$e$$, 10, and 2) for complete reference.
The change of base formula is one of the most practical logarithm properties, enabling computation of logarithms in any base using a standard scientific calculator (which typically only has $$\ln$$ and $$\log_{10}$$). It is essential in computer science (converting between binary and decimal logarithms), information theory, and signal processing.
The change of base formula converts $$\log_a(x)$$ to $$\log_b(x)$$:
$$\log_b(x) = \frac{\ln(x)}{\ln(b)} = \frac{\log_a(x)}{\log_a(b)}$$
Equivalently, the relationship between logarithms in two bases is:
$$\log_b(x) = \log_b(a) \cdot \log_a(x)$$
The conversion factor $$\log_b(a)$$ transforms any base-$$a$$ logarithm into base-$$b$$. For example, to convert from $$\log_{10}$$ to $$\log_2$$, multiply by $$\log_2(10) \approx 3.3219$$.
The calculator computes $$\ln(x)$$, $$\log_{10}(x)$$, and $$\log_2(x)$$ as standard reference values, since these three bases cover most practical applications.
log_a(x) and log_b(x) show the logarithm of your argument in both the original and target bases. The Conversion Factor is the multiplier that converts between bases: $$\log_b(x) = \text{factor} \times \log_a(x)$$. The ln(x), log₁₀(x), and log₂(x) outputs provide the value in all standard bases for quick reference.
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Results
log₁₀(100) = 2. To convert to base 2: log₂(100) = 2 × 3.3219 ≈ 6.644. The conversion factor log₂(10) ≈ 3.3219.
Inputs
Results
log₂(256) = 8. Converting to base 16: log₁₆(256) = 8 × 0.25 = 2 (since 16² = 256).
Most calculators only have $$\ln$$ and $$\log_{10}$$ buttons. The change of base formula lets you compute $$\log_b(x)$$ for any base: $$\log_b(x) = \frac{\ln(x)}{\ln(b)}$$ or $$\frac{\log_{10}(x)}{\log_{10}(b)}$$.
Divide by $$\ln(2)$$: $$\log_2(x) = \frac{\ln(x)}{\ln(2)} = \frac{\ln(x)}{0.6931}$$. Equivalently, multiply $$\ln(x)$$ by $$\frac{1}{\ln(2)} \approx 1.4427$$.
$$\log_{10}(x) = \frac{\ln(x)}{\ln(10)} \approx \frac{\ln(x)}{2.3026}$$. And $$\ln(x) = \log_{10}(x) \times \ln(10) \approx \log_{10}(x) \times 2.3026$$.
Binary systems use base 2. $$\log_2(n)$$ gives the number of bits needed to represent $$n$$ values, the depth of a balanced binary tree with $$n$$ nodes, and the time complexity of binary search ($$O(\log_2 n)$$).
Yes. $$\log_b(x)$$ is defined for any $$b > 0, b \neq 1$$. With $$0 < b < 1$$, the logarithm is negative for arguments greater than 1, since a fraction raised to a positive power is less than 1.
Yes. $$\log_a(b) = \frac{1}{\log_b(a)}$$. This reciprocal relationship follows directly from the change of base formula and is useful for converting between any two bases.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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