Relationship of Logs and AntiLogs 
If f(x) = Log(x), then, f^{ 1}(x) = AntiLog (x). Also, AntiLog (x) = 10^{x}. 
Log(x) = y, 10^{y} = x. 
Log(100) = 2, 10^{2} =100 

AntiLogs 
This column is used to give a detailed explanation of the material. It is my intention that you NOT just read the material but actually get your textbook, notebook/paper, and a pen/pencil and work through the problem with me. Copy down the examples in your note book as you would do in group. When you are done with this page you should be able to do problems 48 in your text in section 7.13.  This column demonstrates the key steps to each example. These key steps also represent the minimum amount of work that needs to be shown as recommended by the Sturgeon General (a mythical fish of the sea that knows a lot about math, human nature, and how your work is graded). 
Example 1 
Example 1: Find AntiLog(2) Solution: All you need to know is that AntiLog(X) = 10^{x}. 
What the problem looks like on paper: AntiLog(2) = 10^{2} =100 
Example 2 
Example 2: Find AntiLog(.2817) Solution: Go to the table in the back of your book and try to find .2817 in the inside part of the table. Go look for it now. You should find a .2810 and a .2833. Since .2817 is closer to .2810, look up the numbers that correspond to the .2810. You should notice the number is 1.9 on the side and the top number is .01. Together they give us a 1.91. Remember you are using a Log table to look up AntiLogs so you do everything backwards compared to looking up Logs. If you got the wrong answer make sure you used the correct table (page 289). 
What the problem looks like on paper: AntiLog(.2817) = 10^{.2817} =1.91 
Example 3 
Example 3: Find AntiLog(4.8779) Solution: Go to the table in the back of your book and try to find 4.8779 in the inside part of the table. Go look for it now. You should notice that the number in the inside are all less than 1.000 so you can't just look up the 4.8779. Instead you have to remember that when you multiply numbers with the same base you add the exponents. Example, (X^{2})(X^{3})=(X^{5}). So, 10^{4.8779}=10^{4}x10^{.8779}. You already know that 10^{4}= 10,000, so just look for the .8779 in the inside of the Log table. This time the .8779 is in the table and the number on the side is 7.5 and the number on the top is .05. Together the give you 7.55. Thus, 10^{4}x10^{.8779} =10,000x10^{.8779} =10,000x7.55 =75,500 Remember you are using a Log table to look up AntiLogs so you do everything backwards compared to looking up Logs. If you got the wrong answer make sure you used the correct table (page 289). 
What the problem looks like on paper: AntiLog(4.8779) = 10^{4.8779} 10^{4.8779}=10^{4}x10^{.8779} =10,000x10^{.8779} =10,000x7.55 =75,500 Warning! The above work represents the minimum amount of work that needs to be shown. Skipping steps and showing any less work could be hazardous to your grade! 
Example 4 
Example 4: Find AntiLog(4.8779) Solution: Go to the table in the back of your book and try to find 4.8779 in the inside part of the table. Go look for it now. You should notice that there are no negative numbers! You also can't just look up 4.8779 and make it negative. Instead you still have to remember that when you multiply numbers with the same base you add the exponents. Example, (X^{2})(X^{3})=(X^{5}). So, 10^{4.8779}=10^{4}x10^{.8779 }???No! =10^{4}x10^{.8779 }Yes! So if the table does not have negative numbers what do you do? You will have to make it positive by doing the following: 10^{4}x10^{.8779}=10^{4}x10^{1}x10^{1}x10^{.8779}, Note that the 10^{1}x10^{1}cancel each other out, 10^{1}x10^{1}= 10^{0}=1. But by grouping the problems as, [10^{4}x10^{1}]x[10^{1}x10^{.8779}], you get, 10^{5}x10^{.1221} Sure you still have the 10^{5}, but 10^{5}= .00001. and 10^{.1221} can be looked up in the Log table now. Again .1221 is not in the table and the numbers nearest to it are .1206 and .1239. Since .1221 is closer to .1206, the corresponding number is 1.32. The rest of the problem looks like, =10^{5}x10^{.1221} =.00001x1.32 =.0000132.
Remember you are using a Log table to look up AntiLogs so you do everything backwards compared to looking up Logs. If you got the wrong answer make sure you used the correct table (page 289). 
What the problem looks like on paper: AntiLog(4.8779) = 10^{4.8779} =10^{4}x10^{.8779} =10^{5}x10^{.1221} =.00001x1.32 =.0000132 Warning! The above work represents the minimum amount of work that needs to be shown. Skipping steps and showing any less work could be hazardous to your grade! 