In C, log10() returns the base-10 logarithm of x. log10(100) is 2, and log10(1) is 0.
You pass one double, and it must be greater than 0. The result is a double. Include <math.h>.
Pass a number greater than 0. Start with this call.
log10(100.0);| Call | What it does |
|---|---|
log10(100.0) | Returns 2, because 10² is 100. |
Include <math.h>. On GCC, link with -lm. For the natural log, use log().
double log10(double x);
float log10f(float x);
long double log10l(long double x);| Parameter | Description |
|---|---|
| x | Required. A double greater than 0. |
| Return Value | Description |
|---|---|
| Base-10 log | A double. log10(100) is 2. |
| NaN | Returned when x is negative. |
Landmark powers of ten return clean integers.
#include <stdio.h>
#include <math.h>
int main(void) {
printf("log10(1) = %.1f\n", log10(1.0));
printf("log10(10) = %.1f\n", log10(10.0));
printf("log10(1000) = %.1f\n", log10(1000.0));
return 0;
}log10.10^0 = 1.10^3).For an integer n > 0, the digit count is floor(log10(n)) + 1.
#include <stdio.h>
#include <math.h>
int main(void) {
int n = 12345;
int digits = (int)floor(log10((double)n)) + 1;
printf("n = %d\n", n);
printf("digits = %d\n", digits);
return 0;
}log10 needs a floating argument.log10(12345) is a bit over 4.floor gives 4; add 1 for five digits.pow()Raising 10 to log10(x) recovers x (within floating-point error).
#include <stdio.h>
#include <math.h>
int main(void) {
double x = 250.0;
double t = log10(x);
double back = pow(10.0, t);
printf("x = %.6f\n", x);
printf("log10(x) = %.6f\n", t);
printf("pow(10, t) = %.6f\n", back);
return 0;
}log10(250) is about 2.398.pow(10.0, t) rebuilds the original scale.log10 + pow(10, …), not exp/log, when thinking in decades.Split a floating value into integer and fractional parts.
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