Question 204023: Using a table of logarithms...how do I find the value of
(.68)to the fifth meaning (.68)exponent 5 ? Found 2 solutions by RAY100, jsmallt9:Answer by RAY100(1637) (Show Source):
You can put this solution on YOUR website! Using a table of logarithms? I can only guess the problem is meant to force you to learn to "fly on manual".
Let's start by writing the problem as an equation, calling the answer we seek "x":
Now we'll find the log of both sides:
Next we'll use one of the properties of logarithms which allows us to "remove" an exponent from the argument: . Using this on our equation to "remove" the 5:
Your table of logarithms probably has columns titled "x" and "log x" or "n" and "log n" or something like that.
If the "x" column has a value for .68 then use the value for "log x" across from it.
If there is no .68 but you have a 6.8 in the "x" column, take the value for "log x" of 6.8 and subtract 1. (This is because .68 = 6.8 x .
If you have a value for 68 (and not .68 or 6.8) in the "x" column, take the value of "log x" for 68 and subtract 2.
One way or another you should get a number that starts with -0.1
Now we will replace with the -0.1... number:
5*(-0.1....) = log(x)
Now multiply the left side getting something that probably starts with
-0.8...:
-0.8... = log(x)
The last part is to find the number whose log is -0.8.... We do this with the following steps:
Since the number is negative, add 1's until it becomes positive. In your case we only need to add 1 once, getting 0.1....
Go to your table and, under the "log x" column, find the closest number to 0.1....
Find the number in the "x" column accross from the 0.1... in the "log x" column. This is your answer!
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