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Showing content from https://en.cppreference.com/w/cpp/language/../algorithm/../../cpp/../c/numeric/math/log1p.html below:

log1p, log1pf, log1pl - cppreference.com

float       log1pf( float arg );

(1) (since C99)

double      log1p( double arg );

(2) (since C99)

long double log1pl( long double arg );

(3) (since C99)

#define log1p( arg )

(4) (since C99) 1-3)

Computes the natural (base

e

) logarithm of

1 + arg

. This function is more precise than the expression

log(1 + arg)

if

arg

is close to zero.

4) Type-generic macro: If arg has type long double, log1pl is called. Otherwise, if arg has integer type or the type double, log1p is called. Otherwise, log1pf is called.

[edit] Parameters arg - floating-point value [edit] Return value

If no errors occur ln(1 + arg) is returned.

If a domain error occurs, an implementation-defined value is returned (NaN where supported).

If a pole error occurs, -HUGE_VAL, -HUGE_VALF, or -HUGE_VALL is returned.

If a range error occurs due to underflow, the correct result (after rounding) is returned.

[edit] Error handling

Errors are reported as specified in math_errhandling.

Domain error occurs if arg is less than -1.

Pole error may occur if arg is -1.

If the implementation supports IEEE floating-point arithmetic (IEC 60559),

[edit] Notes

The functions expm1 and log1p are useful for financial calculations, for example, when calculating small daily interest rates: (1+x)n
-1
can be expressed as expm1(n * log1p(x)). These functions also simplify writing accurate inverse hyperbolic functions.

[edit] Example
#include <errno.h>
#include <fenv.h>
#include <float.h>
#include <math.h>
#include <stdio.h>
// #pragma STDC FENV_ACCESS ON
 
int main(void)
{
    printf("log1p(0) = %f\n", log1p(0));
    printf("Interest earned in 2 days on $100, compounded daily at 1%%\n"
           " on a 30/360 calendar = %f\n",
           100*expm1(2*log1p(0.01/360)));
    printf("log(1+1e-16) = %g, but log1p(1e-16) = %g\n",
           log(1+1e-16), log1p(1e-16));
 
    // special values
    printf("log1p(-0) = %f\n", log1p(-0.0));
    printf("log1p(+Inf) = %f\n", log1p(INFINITY));
 
    // error handling
    errno = 0; feclearexcept(FE_ALL_EXCEPT);
    printf("log1p(-1) = %f\n", log1p(-1));
    if (errno == ERANGE)
        perror("    errno == ERANGE");
    if (fetestexcept(FE_DIVBYZERO))
        puts("    FE_DIVBYZERO raised");
}

Possible output:

log1p(0) = 0.000000
Interest earned in 2 days on $100, compounded daily at 1%
 on a 30/360 calendar = 0.005556
log(1+1e-16) = 0, but log1p(1e-16) = 1e-16
log1p(-0) = -0.000000
log1p(+Inf) = Inf
log1p(-1) = -Inf
    errno == ERANGE: Result too large
    FE_DIVBYZERO raised
[edit] References
[edit] See also

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