Average Error: 5.7 → 0.4
Time: 2.8s
Precision: binary64
\[e^{\left(1 + \log a\right) + \log b}\]
↓
\[\left(b \cdot e^{1}\right) \cdot a\]
e^{\left(1 + \log a\right) + \log b}
↓
\left(b \cdot e^{1}\right) \cdot a
double code(double a, double b) {
	return ((double) exp(((double) (((double) (1.0 + ((double) log(a)))) + ((double) log(b))))));
}
↓
double code(double a, double b) {
	return ((double) (((double) (b * ((double) exp(1.0)))) * a));
}

Error

Bits error versus a

Bits error versus b

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 5.7

    \[e^{\left(1 + \log a\right) + \log b}\]

  2. Simplified0.4

    \[\leadsto \color{blue}{\left(b \cdot e^{1}\right) \cdot a}\]

  3. Final simplification0.4

    \[\leadsto \left(b \cdot e^{1}\right) \cdot a\]

Reproduce

herbie shell --seed 2020153 
(FPCore (a b)
  :name "(exp (+ (+ 1 (log a)) (log b)))"
  :precision binary64
  (exp (+ (+ 1.0 (log a)) (log b))))