not the same solution as this

In a function, I need to calculate the exponential of a fraction. The exponent is a whole number, and the base is not.

I tried using the ABDKMath64x64 library and using this formula: enter image description here

I converted r into a 64.64bit fixed float which is this: (1.8538977)×10^19÷2^64 ≈ 1.0049999569529399767731092651956714689731597900390625

(My r value is 0.005 or 0.5%)

And I got the majority of my code working here

    int128 nominator = 1.8538977 * 1e19;
    nominator = ABDKMath64x64.log_2(nominator);
    int128 newNominator = nominator * rebaseAmount;
    int128 result = ABDKMath64x64.exp_2(newNominator); 
    uint64 uintResult = ABDKMath64x64.toUInt(result);
    return uintResult; 

The problem is the last line. I am certain my int128 result is accurate (when I manually divide it by 2^64, I get my desired number), but when I cast it to a uint, it gives me a single whole number, which is like the current whole number, but rounded down to classic uint style.

Is there a way to cast the result into a uint whilst keeping the following decimals? I tried doing * 1e18 to the above code but the decimals don't keep due to the conversion.

If not, are there better ways at solving 1.005^x, than using this library and formula?

1 Answer 1


Wow, I solved this by just adding 1e18 to the int128 result line, and then doing the function. I thought having a int128 with 36 leading decimals would overflow, but it appears now.

The working code is

        nominator = ABDKMath64x64.log_2(nominator);
        int128 newNominator = nominator * rebaseAmount;
        int128 result = ABDKMath64x64.exp_2(newNominator) * 1e18 ; 
        uint64 resultUint = ABDKMath64x64.toUInt(result);
        return resultUint;

However, I am limited as rebaseAmount overflows for an amount ~500. Are there other implementations possible here?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.