SOLUTION: To four decimal places, the values of log2 and log3 are Log2 = 0.3010 and log3 = 0.4771 Evaluate log (1/9).

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: To four decimal places, the values of log2 and log3 are Log2 = 0.3010 and log3 = 0.4771 Evaluate log (1/9).      Log On


   

Question 814157: To four decimal places, the values of log2 and log3 are
Log2 = 0.3010 and log3 = 0.4771
Evaluate log (1/9).
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Here's a procedure for solving these types of problems where they give you the value of some logarithms and ask you to find the value of another log (without using a calculator):
  1. Rewrite the argument of the desired log as a product, quotient and/or powers of the numbers whose logs you know. Note: In addition to the logs you're given you should also know that log%28a%2C+%28a%29%29+=+1. So, since we're using base 10 logs in this problem, we should know that log(10) = 1 without it being given to you. So in this problem we start knowing the base 10 logs of 2, 3 and 10.
  2. Use properties of logarithms to rewrite the expression in terms of log's of 2's, 3's and/or 10's.
  3. Substitute in the known values for these logs and simplify.
So for the first step we will want to rewrite 1/9 as a product, quotient and/or power of 2's, 3's and/or 10's. I hope it doesn't take long for you to see that 9 is a power of 3. But what about 1/9? If you know your negative exponents then you will see that 1%2F9+=+3%5E%28-2%29. So
log%28%281%2F9%29%29
can be rewritten as:
log%28%283%5E%28-2%29%29%29
Now we need to rewrite this so that it is in terms of log(3). So the -2 has to move. For this we use a property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29
%28-2%29log%28%283%29%29
And finally we substitute in the known value for log(3):
%28-2%29%2A+0.4771
which simplifies to:
-0.9542