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:\chap11>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 IO.
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 Java documentation you can download from java.sun.com. If you don't have this on your computer, you can search for "sqrt Java" with a search service (like Google) and find it.
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?