Only two lines need to be changed:
double result = Math.log( value );
System.out.println("logarithm: " + result );
sqrt() MethodHere is sample output from the (unmodified) program:
C:\chap13>java SquareRoot Enter a double: 3 square root : 1.7320508075688772
It is OK to enter a number without a decimal point
(as the "3" above) for floating point I/O.
Java converts it to the correct type.
Here is some of the documentation for the method sqrt():
static double sqrt(double a) Returns the correctly rounded positive square root of a double value.
This was copied from the on-line Java documentation. Search for "sqrt Java" with a search service (like Google) when you need it.
https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html.
Inspect the documentation.
What is the type of the argument expected by sqrt()?
What is the type of value returned by sqrt()?
Is sqrt() a static method?