user: Demonstrate the C code that generates the assembly output below:

```asm
<Logarithm>:
  endbr64
  push   %r12
  lea    D31(%rip),%rdi
  push   %rbp
  lea    D63(%rip),%rbp
  sub    $0x28,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  lea    0xc(%rsp),%r12
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    $0x1,%eax
  je     L43
  cmp    $0x2,%eax
  je     L5
  cmp    $0x3,%eax
  je     L28
L47:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L3
  add    $0x28,%rsp
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L43:
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%edi
  call   <log10@plt>
  mov    0xc(%rsp),%esi
  mov    $0x1,%eax
  lea    D55(%rip),%rdi
  call   <printf@plt>
  jmp    L47
  nopl   0x0(%rax,%rax,1)
L5:
  lea    D40(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%ebp
  pxor   %xmm0,%xmm0
  cvtsi2sd %ebp,%xmm0
  call   <log@plt>
  mov    %ebp,%esi
  mov    $0x1,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L47
  nopl   0x0(%rax)
L28:
  lea    D48(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r12,%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  lea    D54(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x10(%rsp),%rsi
  mov    %rbp,%rdi
  xor    %eax,%eax
  call   <scanf@plt>
  mov    0xc(%rsp),%r12d
  pxor   %xmm0,%xmm0
  cvtsi2sd %r12d,%xmm0
  call   <log@plt>
  movq   %xmm0,%rbp
  movsd  0x10(%rsp),%xmm0
  call   <log@plt>
  movapd %xmm0,%xmm1
  movq   %rbp,%xmm0
  divsd  %xmm1,%xmm0
  call   <log@plt>
  mov    %r12d,%esi
  mov    $0x1,%eax
  lea    D3(%rip),%rdi
  call   <printf@plt>
  jmp    L47
L3:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D31
data_type: string
name: parse_data
arguments: tag: D63
data_type: string
name: parse_data
arguments: tag: D40
data_type: string
name: parse_data
arguments: tag: D55
data_type: string
name: parse_data
arguments: tag: D3
data_type: string
name: parse_data
arguments: tag: D48
data_type: string
name: parse_data
arguments: tag: D54
data_type: string

tool: D31: 1-Compute log base 10 
2- calculate natural logarithm 
3-Custom Log 

 Enter Choice
>>

tool: D63: %d

tool: D40: Enter Number

>>

tool: D55: Logarithm %d to the base 10 = %f

tool: D3: Logarithm %d to the base e = %f

tool: D48: Enter base

>>

tool: D54: Enter number

>>

assistant: ```c
void Logarithm() {
  int input, output;
  double i, x;
  printf("1-Compute log base 10 \n2- calculate natural logarithm \n3-Custom Log \n\n Enter Choice\n>>");
  scanf("%d", &input);
  if (input == 1) {
    printf("Enter Number\n\n>>");
    scanf("%d", &input);
    printf("Logarithm %d to the base 10 = %f", input, log10(input));
  } else if (input == 2) {
    printf("Enter Number\n\n>>");
    scanf("%d", &input);
    printf("Logarithm %d to the base e = %f", input, log(input));
  } else if (input == 3) {
    printf("Enter base\n\n>>");
    scanf("%d", &input);
    printf("Enter number\n\n>>");
    scanf("%d", &x);
    i = log(input) / log(x);
    printf("Logarithm %d to the base e = %f", input, log(i));
  }
}
```