How to calculate the logarithm of a number in MASM 32

How do I calculate the logarithm of a number in MASM 32?
For example if I have to calculate log(2.5), how will I do this?
I know this will involve fyl2x and I have tried but I couldn't calculate it accurately.

This is what I tried but it prints no result.

INCLUDE Irvine32.inc

; .data is used for declaring and defining variables
.data
num     real8 3.324          ; the data I want to convert
res     real8 ?              

   .code
   main PROC

    fldl2t               ; st: log2(10)
    fld num              ; st: log2(10) num
    fyl2x                ; st: log10(num)
    fstp res             ; store to out and pop


    call    CrLf
    call    CrLf

exit    
main ENDP
END main

I see that you have copied the code from an existing answer. Sadly, that answer contains errors!

Before using the FYL2X instruction, read the description in the manual:

This instruction calculates (ST(1)*log₂(ST(0))), stores the result in register ST(1), and pops the FPU register stack. The source operand in ST(0) must be a non-zero positive number.

The FYL2X instruction is designed with a built-in multiplication to optimize the calculation of logarithms with an arbitrary positive base (b).
log_bx = (log₂b)^-1 * log₂x

The error stems from not noticing that the multiplier is a reciprocal. We can easily remove it because (log₂b)^-1 = log_b2 and then we substitute 10 for b. We therefore need as multiplier log₁₀2 which the FPU provides through its FLDLG2 instruction.

num     REAL8 3.324
res     REAL8 ?

        ...

        FLDLG2           ; ST0 = log10(2)
        FLD     num      ; ST1 = log10(2), ST0 = num
        FYL2X            ; ST0 = log10(num)
        FSTP    res

To calculate the natural logarithm (aka 'ln'), the only change would be to use FLDLN2 (instead of FLDLG2) to obtain the multiplier log_e2.

To calculate the binary logarithm, the only change would be to use FLD1 (instead of FLDLG2) to obtain the trivial multiplier 1.

An algorithm that converts a floating point number into its textual representation is not all that simple, but maybe next fixed-point trick is all that you need:

• multiply the floating point number by 1000 (for a 3-digit precision)
• store the product to memory as integer
• convert (*) the integer into text (repeated divisions by 10)
• put a decimal point character between the last 3 digits and all the digits in front of them

num     REAL8 3.324
quad    dq ?
kilo    dw 1000

        ...

        FLDLG2           ; ST0 = log10(2)
        FLD     num      ; ST1 = log10(2), ST0 = num
        FYL2X            ; ST0 = log10(num)
        FIMUL   kilo     ; ST0 = 1000 * log10(num)
        FISTP   quad
        ...

(*) Displaying numbers with DOS explains how converting an integer to text works. Look through the 'with DOS' mention and focus on the how and the why of the conversion itself.