SOLUTION: Describe how to obtain the third answer in each row from the first two questions? a) log [base 8]512, log [base 2] 512,log [base 16] 512 b) log [base 4]64, log [base 8]64, log

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Describe how to obtain the third answer in each row from the first two questions? a) log [base 8]512, log [base 2] 512,log [base 16] 512 b) log [base 4]64, log [base 8]64, log       Log On


   

Question 155459: Describe how to obtain the third answer in each row from the first two questions?
a) log [base 8]512, log [base 2] 512,log [base 16] 512
b) log [base 4]64, log [base 8]64, log [base 32]64
I don't know how they got the third answer for these two sequences. Can you show the steps on how you got the third answer? This way I can understand. Thank you for all the hard work!! Also, I don't know which ISBN to put because my question isn't from a textbook. Sorry!
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
First, you recall that the definition of the logarithm of a numer is:
"The logarithm of a number is the power to which the base of the logarithm must be raised to equal the number"
a) Log%5B16%5D%28512%29+=+x so, by the above definition, x is the power to which 16 (the base) must be raised to equal the number (512). So you can write that out as described:
16%5Ex+=+512 The next step is to exprees the the numbers on both sides as powers of the same base.
16+=+2%5E4 and 512+=+2%5E9 so make the appropriate substitutions to get:
2%5E%284x%29+=+2%5E9 now since the bases (2) are equal, the exponents are equal, so...
4x+=+9 and
x+=+9%2F4 so
Log%5B16%5D%28512%29+=+9%2F4 or
Log%5B16%5D%28512%29+=+2.25
Similarly:
b) Log%5B32%5D%2864%29+=+x
32%5Ex+=+64 but32+=+2%5E5 and 64+=+2%5E6 so...
2%5E5x+=+2%5E6 so...
5x+=+6 and..
x+=+6%2F5 then...
Log%5B16%5D%2864%29+=+6%2F5 or
Log%5B16%5D%2864%29+=+1.2