How to find every integer that is the closest to the antilogarithm (10base) for a series of n*0.1 (in Python)
Question: In my algorithm I want to evaluate if the current integer n happens to be
closest to the antilogarithm of any positive multiple of 0.1, so 0.1, 1.0,
1.1, 7.9, 21.5 etc.

Antilog? AntiLog(x) = 10x

I found an online calculator that finds the antilog here:
<http://ncalculators.com/number-conversion/anti-log-logarithm-calculator.htm>
but nowhere I could find any examples of doing this in Python, or any other
programming language.

If I can't use Python to find the antilog of a series of numbers I would have
to resort to storing a list of antilog values in my program, for performance
I'm might consider that anyways but nevertheless it would be great to figure
out how to do this in code.

**Update:** With the code from first answer I was able to do it, this code
fully demonstrates all I was trying to do:

    
    
    #!/usr/bin/python
    import sys
    import math
    
    for x in xrange(1, 1000000):
        target = round(math.log10(x),1)
        int_antilog = int(10**target+0.5) #fast round() alternative
        if (x == int_antilog):
            print 'do something at call ' + str(x)
    


Answer: Python has a [power
operator](http://docs.python.org/2/reference/expressions.html#the-power-
operator) that performs the exponentiation you want to do:

    
    
    def antilog(x):
        return 10 ** x
    

Examples:

    
    
    >>> antilog(0.1)
    1.2589254117941673
    >>> antilog(3)
    1000
    >>> inputs = [0.1, 1.0, 1.1, 7.9, 21.5]
    >>> outputs = [antilog(x) for x in inputs]
    >>> print outputs
    [1.2589254117941673, 10.0, 12.589254117941675, 79432823.47242822, 3.1622776601683794e+21]
    

Round and convert to integers in your favourite way, and you'll be good to go.

