SOLUTION: Given that log 2 ≈ 0.301 and log 3 ≈ 0.477, find the following. log6 27 Here is the work I have done, can someone check it for me? log(6) = log(2*3) log(2) lo

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Given that log 2 ≈ 0.301 and log 3 ≈ 0.477, find the following. log6 27 Here is the work I have done, can someone check it for me? log(6) = log(2*3) log(2) lo      Log On


   

Question 270469: Given that log 2 ≈ 0.301 and log 3 ≈ 0.477, find the following.
log6 27
Here is the work I have done, can someone check it for me?
log(6) = log(2*3)
log(2) log(3) = .301 .477 = .778
log6(27) = 1.84
Answer by jsmallt9(3759) About Me  (Show Source):
You can put this solution on YOUR website!
log%286%2C+%2827%29%29
The key to this problem is to realize that a base 6 logarithm is hard to work with. We need to change the base to one that is easy to work with, like base 10 logarithms. Fortunately there is a formula for doing just this: log%28a%2C+%28x%29%29+=+log%28b%2C+%28x%29%29%2Flog%28b%2C+%28a%29%29. Using this formula to change your base 6 logarithm to an expression of base 10 logarithms we get:
log%286%2C+%2827%29%29+=+log%28%2827%29%29%2Flog%28%286%29%29
We now have base 10 logarithms to work with. Next, since we are given log(2) and log(3), we want to express the arguments in terms of 2's and/or 3's. You already figured out how to express a 6 in terms of 2's and 3's. With a little thought I hope you see that 27+=+3%5E3. So rewriting the logarithms using 2's and 3's we get:

Next we can use one property of logarithms, log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29 to move the exponent of 3%5E3 out in front of the logarithm. And we can use another property of logarithms, log%28a%2C+%28p%2Aq%29%29+=+log%28a%2C+%28p%29%29+%2B+log%28a%2C+%28b%29%29, to split the log(2*3) into separate logarithms:

We have finally changed our original base 6 logarithm of 27 into an expression of base 10 logarithms of 2 and 3. We can now use the given values for log(2) and log(3) and simplify:

Rounded to the nearest 1/100 this matches the answer you got. But I have no idea how you came up with this. The work you provided is grossly incomplete.