Instantly convert a logarithm from one base to another using the change of base formula. Works with any positive base (except 1). You can even use e for natural log.
e for Euler's number (natural base) if needed.
⚡ The result means:
\( \log_{b}(x) = K \cdot \log_{B}(x) \) with \( K = \frac{1}{\log_B(b)} \).
The calculator shows \( K \) directly.
Given \( \log_b (x) \), you can change to any base \( B \) ( \( B>0, B\neq 1 \) ) :
\[ \log_b (x) = \frac{\log_B (x)}{\log_B (b)} \]
Equivalently: \(\displaystyle \log_b (x) = \left(\frac{1}{\log_B(b)}\right) \log_B(x)\)
Change \( \log_2 (x) \) to natural base \( e \).
\( b = 2,\; B = e \)
\[ \log_2(x) = \frac{\log_e(x)}{\log_e(2)} = \frac{1}{\ln 2}\,\ln x\]
\(\displaystyle \frac{1}{\ln 2} \approx 1.4427\) → \(\log_2(x) \approx 1.4427\,\ln x\)
Rewrite \( \log_4 (x) \) with base \( 2 \).
\( b = 4,\; B = 2 \)
\[ \log_4(x) = \frac{\log_2(x)}{\log_2(4)} = \frac{\log_2(x)}{2} = 0.5\,\log_2(x)\]
So \(\log_4(x) = 0.5 \log_2(x)\)
Try these manually, then check with the calculator above. Enter \(b\) (original base) and \(B\) (new base).