How to use the IMLOG2 function
The IMLOG2 function calculates the base 2 logarithm of a complex number in x + yi or x + yj text format.
The letter j is used in electrical engineering to distinguish between the imaginary value and the electric current.
Provide an example equation of when the base 2 logarithm is needed with real values ?
This equation can be solved using the base 2 logarithm: 2x=16
becomes
log2(2x) = log2(16)
becomes
x log2(2) = 4
log2(2) = 1
x*1=4
equals
x = 4
What is the difference between the natural logarithm and the base 2 logarithm?
The natural logarithm uses e as the base and the log2 uses 2 as the base.
What is the difference between the IMLOG2 function and the IMLOG10 function?
The IMLOG2 function uses 2 as the base and the IMLOG10 function uses 10 as the base.
Table of Contents
1. IMLOG2 Function Syntax
IMLOG2(inumber)
2. IMLOG2 Function Arguments
| inumber | Required. A complex number in x+yi or x+yj text format. |
3. IMLOG2 function example
The image above demonstrates a formula in cell B28 that calculates the base 2 logarithm of a complex number specified in cell B25.
Cell C28 calculates the real number from the complex number in cell B28. Cell D28 extracts the imaginary number from the complex number specified in cell B28.
The real and imaginary numbers separated in a cell each allow us to graph the complex number on the complex plane.
Formula in cell B28:
The chart above shows the complex plane, the y-axis is the imaginary axis and the x-axis is the real axis.
Complex number 5+5i is the light blue line in the first quadrant. The base 2 logarithm of 5+5i is the green line also in the first quadrant.
3.1 Explaining formula
Step 1 - Populate arguments
IMLOG2(inumber)
becomes
IMLOG2(B25)
Step 2 - Evaluate the IMLOG2 function
IMLOG2(B25)
becomes
IMLOG2("5+5i")
and returns
2.82192809488736+1.1330900354568i
4. How to calculate the base 2 logarithm of a complex number in detail?
The base 2 logarithm of a complex number is calculated like this:
C = x + yi
IMLOG2(C) =log2(e)ln(x+yi)
For example, C = 5 + 5i
IMLOG2(C) =log2(e)ln(x+yi) = log2(e)ln(5+5i)
IMLOG2(C) = log2(e)ln(5+5i) = 1.44269504088896*ln(5+5i)
IMLOG2(C) = 1.44269504088896*ln(5+5i) =1.44269504088896*(1.95601150271407+0.785398163397448i)
IMLOG2(C) = 1.44269504088896*(1.95601150271407+0.785398163397448i) = 2.82192809488736+1.1330900354568i
5. The IMLOG2 function not working - #NUM error
The base 2 logarithm function IMLOG2(x+yi) is not defined for x=0, IMLOG2(0) is not a valid expression.
The IFERROR function can help you handle errors by returning a value of your choice if an error value occurs.
Why is it not possible to calculate the base 10 logarithm of a complex number that has a real part of zero and an imaginary part of 0?
The formula to calculate the base 2 logarithm of a complex number is log10(x+yi)=(log10e)ln(x+yi)
There is no solution to ln(0) which is why the calculation is not possible.
Useful links
IMLOG2 function - Microsoft
Binary logarithm
Log base 2
Functions in 'Engineering' category
The IMLOG2 function function is one of 42 functions in the 'Engineering' category.
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