Sig Fig for Logs Calculator
Use this free sig fig for logs calculator to find the antilog (10^x or e^x) of a logarithmic value with the correct number of significant figures. Enter a log value and get the result with proper sig fig handling.
Enter Values
The logarithmic value to take the antilog of
Choose antilog base 10 or base e
Result
Enter values above and click Calculate to see your result.
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Formula
When computing an antilog (10^x or e^x), the number of decimal places in the logarithmic value determines the number of significant figures in the result.
Worked Example
Sig Fig Rules for Antilogs
- If pH = 3.42 (2 decimal places), then [H+] = 10^-3.42 = 3.8 x 10^-4 (2 sig figs)
- The characteristic (integer part) of the log only sets the order of magnitude
- Only the mantissa (decimal part) determines precision in the result
- This applies to both base 10 (antilog) and base e (e^x) calculations
Antilog sig fig rules are critical in chemistry for converting pH to hydrogen ion concentration, pKa to Ka values, and any situation where you reverse a logarithmic transformation.
You can also calculate changes using our Log Sig Fig Calculator, Sig Fig Rules Calculator or Chemistry Sig Fig Calculator.
Frequently Asked Questions
What is an antilog?
An antilog is the reverse of a logarithm. If log(x) = y, then the antilog of y is x. For base 10, antilog(y) = 10^y. For natural log, antilog(y) = e^y.
How do decimal places in a log become sig figs?
The integer part of a log (characteristic) only indicates the power of 10. The decimal part (mantissa) carries precision. So when you reverse the operation, only the mantissa-derived precision transfers to the result as sig figs.
How do I convert pH back to concentration?
[H+] = 10^(-pH). If pH = 4.75 (2 decimal places), then [H+] = 10^-4.75 = 1.8 x 10^-5, reported with 2 sig figs.
What if my log has no decimal places?
A log value with no decimal places (like 3) implies minimal precision. The antilog (10^3 = 1000) would have at best 1 sig fig, though in practice such a value is often treated as exact depending on context.
Does this work for pKa to Ka conversions?
Yes. Ka = 10^(-pKa). If pKa = 4.76 (2 decimal places), then Ka = 10^-4.76 = 1.7 x 10^-5 (2 sig figs).
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