can you please modify the code to calculate the exponent of a nember instead of incorrectly multiply:.ORIG x3000 ; Clear registers AND R0, R0, #0 AND R1, R1, #0 AND R2, R2, #0 ; Load x and y variables LD R0, VAR_X LD R1, VAR_Y ; Loop 1 iterates y times Loop1 ADD R2, R2, R0 ; Multiply R2 by R0 ; Loop 1 end ADD R1, R1, #-1 BRn EndLoop1 ; Loop 2 multiplies R0 by x Loop2 ADD R0, R0, #-1 ; Loop 2 end BRp Loop1 EndLoop1 ; Answer in R2 ; Store the result in memory address x8000 STI R2, PTR_OUT ; Halt the program HALT VAR_X .FILL x0003 VAR_Y .FILL x0004 PTR_OUT .FILL x8000 .END ### Introduction to Exponential Operations in Assembly LanguageIn this exercise, we will focus on writing a program that calculates the result of an exponential operation, specifically \( z = x^y \), using assembly language. The steps outlined below will guide you in completing this task, ensuring that variables are correctly stored and utilized in memory, and that the result is correctly stored at a specified memory address.#### Instructions1. **Write a program that calculates the result of \( z = x^y \)** a. **Memory Initialization**: - The variables \( x \) and \( y \) should be loaded into memory before the calculation begins. This can be achieved by using the `.FILL` pseudo-instruction to initialize these variables in memory. b. **Variable Constraints**: - Ensure that the variables \( x \) and \( y \) fall within the specified range: - \( 0 \leq x \leq 6 \) - \( 0 \leq y \leq 6 \) c. **Result Storage and Program Termination**: - Once the exponential operation is performed, the result should be stored in the memory address `x8000`. - The program should then halt execution. #### Example CodeBelow is a conceptual example (the syntax will vary depending on the specific assembly language you are using):```assembly.ORIG x3000 ; Starting addressLD R0, X ; Load x into register R0LD R1, Y ; Load y into register R1 EXPONENT R2, R0, R1 ; Perform exponentiation x^y, store result in R2ST R2, RESULT ; Store result in memory address x8000HALT ; Halt programX .FILL x5 ; Example value for x (in hex)Y .FILL x3 ; Example value for y (in hex)RESULT .FILL x8000 ; Address for the result.END```#### Explanation- **.ORIG x3000**: This directive sets the starting address of the program to `x3000`.- **LD R0, X**: Loads the value of `X` into register `R0`.- **LD R1, Y**: Loads the value of `Y` into register `R1`.- **EXP

Dear Student,The program below is written in LC3 assembly language and it is used to calculate the…

Answered: can you please modify the code to…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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can you please modify the code to calculate the exponent of a nember instead of incorrectly multiply:

.ORIG x3000 ; Clear registers AND R0, R0, #0 AND R1, R1, #0 AND R2, R2, #0 ; Load x and y variables LD R0, VAR_X LD R1, VAR_Y ; Loop 1 iterates y times Loop1 ADD R2, R2, R0 ; Multiply R2 by R0 ; Loop 1 end ADD R1, R1, #-1 BRn EndLoop1 ; Loop 2 multiplies R0 by x Loop2 ADD R0, R0, #-1 ; Loop 2 end BRp Loop1 EndLoop1 ; Answer in R2 ; Store the result in memory address x8000 STI R2, PTR_OUT ; Halt the program HALT VAR_X .FILL x0003 VAR_Y .FILL x0004 PTR_OUT .FILL x8000 .END