Hello Subash!
Here is the solution for the question you are asking for, I solved it in procedural way but if you are among the one who prefer OOP style then you can still ask it for me cause I have solved it from both methods but here I am just going to leave procedural one....
//author:Manish Acharya
import java.util.Scanner;
import java.util.*;
public class idgenerator {
public static void main(String[] args) {
String small_name="", long_name="", new_small_name="", new_long_name="";
char lr='a',...
#include<stdio.h>
#include<conio.h>
#include<string.h>
int main() {
int i, number, num1, num2=0;
char str1[50];
num1 = number;
for(i=2; i<=20; i+=2) {
printf("%d", i);
printf(", ");
}
return 0;
}
1. sol:
P | q | ~p | ~p ^ q |
| T T F F | T F T F | F F T T | F F T F |
Log2aa=x then, a=(2a)x ......(1)
Log3a2a=y then,2a=(3a)y ......(2)
Log4a 3a=z then, 3a=(4a)z ......(3)
So,
a=(2a)x [from (1)]
Or, a=(3a)xy [from(2)]
Or, a=(4a)xyz [from(3)]
Multiplying both sides by 4a,
4a.a=4a.(4a)xyz
Or,(2a)² =(4a)xyz + 1
Or,(3a)2y =(4a)xyz+1
Or,(4a)2yz =(4a)xyz+1
Or, 2yz = xyz+1 .proved.

